$f: G to mathbb{C}^*$ is a homomorphism. Show that the sum $sum f (g) = 0$ or $n$












5














Let $ mathbb{C}^*$ be the multiplicative group of non-zero complex numbers. Let $G$ be an abelian group and suppose $f: G to mathbb{C}^*$ is a homomorphism. Prove that $sum_{g in G} f(g)=n$ or, $sum_{g in G} f(g)=0$, where $n =o(G)$



Proof attempt:



The case is evident for the trivial homomorphism; the sum adds up to $n$.



For the second part



We know, the only elements with finite order in the group $ mathbb{C}^*$ are $1$ and $-1$, with $o(-1)=2$.



Now, the only case when $f(g)$ can take $-1$ as a value is when $n$ is even.



Consider the subgroup $({1, -1}, .) = G'$ of the group $ mathbb{C}^*$. We have, from the Isomorphism Theorem, $ G/ ker( f ) simeq G' $ [since $f$ takes each value from $G'$].



As $o(G')=2$, $o(G/ ker( f ))=2$, i.e $o(ker (f))= n/2$. Hence, when summed, the resultant is $0$.



Edit: A foolish assumption has been taken. The finite ordered complex numbers in the said group is of the form $z^n=1$, so I have 'proved' a very restricted case, which is not at all desired.






abstract-algebra group-theory proof-verification finite-groups group-isomorphism







share|cite|improve this question




















  • 6




    What's the order of $i$ then?
    – the_fox
    Jan 4 at 3:11






  • 2




    Does G have to have finite order?
    – Joel Pereira
    Jan 4 at 3:39










  • @JoelPereira:. isn't that implied by the statements $sum f(g) = n$ and $n = o(G)$?
    – Robert Lewis
    Jan 4 at 3:45










  • @the_fox :( back to square one.
    – Subhasis Biswas
    Jan 4 at 11:41






  • 1




    @RobertLewis no it's not implied. If G = the multiplicative group of $mathbb{R}^+$, we can still form the sum. In that case the sum would diverge.
    – Joel Pereira
    Jan 4 at 15:33
















5














Let $ mathbb{C}^*$ be the multiplicative group of non-zero complex numbers. Let $G$ be an abelian group and suppose $f: G to mathbb{C}^*$ is a homomorphism. Prove that $sum_{g in G} f(g)=n$ or, $sum_{g in G} f(g)=0$, where $n =o(G)$



Proof attempt:



The case is evident for the trivial homomorphism; the sum adds up to $n$.



For the second part



We know, the only elements with finite order in the group $ mathbb{C}^*$ are $1$ and $-1$, with $o(-1)=2$.



Now, the only case when $f(g)$ can take $-1$ as a value is when $n$ is even.



Consider the subgroup $({1, -1}, .) = G'$ of the group $ mathbb{C}^*$. We have, from the Isomorphism Theorem, $ G/ ker( f ) simeq G' $ [since $f$ takes each value from $G'$].



As $o(G')=2$, $o(G/ ker( f ))=2$, i.e $o(ker (f))= n/2$. Hence, when summed, the resultant is $0$.



Edit: A foolish assumption has been taken. The finite ordered complex numbers in the said group is of the form $z^n=1$, so I have 'proved' a very restricted case, which is not at all desired.






abstract-algebra group-theory proof-verification finite-groups group-isomorphism







share|cite|improve this question




















  • 6




    What's the order of $i$ then?
    – the_fox
    Jan 4 at 3:11






  • 2




    Does G have to have finite order?
    – Joel Pereira
    Jan 4 at 3:39










  • @JoelPereira:. isn't that implied by the statements $sum f(g) = n$ and $n = o(G)$?
    – Robert Lewis
    Jan 4 at 3:45










  • @the_fox :( back to square one.
    – Subhasis Biswas
    Jan 4 at 11:41






  • 1




    @RobertLewis no it's not implied. If G = the multiplicative group of $mathbb{R}^+$, we can still form the sum. In that case the sum would diverge.
    – Joel Pereira
    Jan 4 at 15:33














5












5








5


1





Let $ mathbb{C}^*$ be the multiplicative group of non-zero complex numbers. Let $G$ be an abelian group and suppose $f: G to mathbb{C}^*$ is a homomorphism. Prove that $sum_{g in G} f(g)=n$ or, $sum_{g in G} f(g)=0$, where $n =o(G)$



Proof attempt:



The case is evident for the trivial homomorphism; the sum adds up to $n$.



For the second part



We know, the only elements with finite order in the group $ mathbb{C}^*$ are $1$ and $-1$, with $o(-1)=2$.



Now, the only case when $f(g)$ can take $-1$ as a value is when $n$ is even.



Consider the subgroup $({1, -1}, .) = G'$ of the group $ mathbb{C}^*$. We have, from the Isomorphism Theorem, $ G/ ker( f ) simeq G' $ [since $f$ takes each value from $G'$].



As $o(G')=2$, $o(G/ ker( f ))=2$, i.e $o(ker (f))= n/2$. Hence, when summed, the resultant is $0$.



Edit: A foolish assumption has been taken. The finite ordered complex numbers in the said group is of the form $z^n=1$, so I have 'proved' a very restricted case, which is not at all desired.






abstract-algebra group-theory proof-verification finite-groups group-isomorphism







share|cite|improve this question















Let $ mathbb{C}^*$ be the multiplicative group of non-zero complex numbers. Let $G$ be an abelian group and suppose $f: G to mathbb{C}^*$ is a homomorphism. Prove that $sum_{g in G} f(g)=n$ or, $sum_{g in G} f(g)=0$, where $n =o(G)$



Proof attempt:



The case is evident for the trivial homomorphism; the sum adds up to $n$.



For the second part



We know, the only elements with finite order in the group $ mathbb{C}^*$ are $1$ and $-1$, with $o(-1)=2$.



Now, the only case when $f(g)$ can take $-1$ as a value is when $n$ is even.



Consider the subgroup $({1, -1}, .) = G'$ of the group $ mathbb{C}^*$. We have, from the Isomorphism Theorem, $ G/ ker( f ) simeq G' $ [since $f$ takes each value from $G'$].



As $o(G')=2$, $o(G/ ker( f ))=2$, i.e $o(ker (f))= n/2$. Hence, when summed, the resultant is $0$.



Edit: A foolish assumption has been taken. The finite ordered complex numbers in the said group is of the form $z^n=1$, so I have 'proved' a very restricted case, which is not at all desired.






abstract-algebra group-theory proof-verification finite-groups group-isomorphism




abstract-algebra group-theory proof-verification finite-groups group-isomorphism






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 11:44







Subhasis Biswas

















asked Jan 4 at 2:51









Subhasis BiswasSubhasis Biswas

441211




441211








  • 6




    What's the order of $i$ then?
    – the_fox
    Jan 4 at 3:11






  • 2




    Does G have to have finite order?
    – Joel Pereira
    Jan 4 at 3:39










  • @JoelPereira:. isn't that implied by the statements $sum f(g) = n$ and $n = o(G)$?
    – Robert Lewis
    Jan 4 at 3:45










  • @the_fox :( back to square one.
    – Subhasis Biswas
    Jan 4 at 11:41






  • 1




    @RobertLewis no it's not implied. If G = the multiplicative group of $mathbb{R}^+$, we can still form the sum. In that case the sum would diverge.
    – Joel Pereira
    Jan 4 at 15:33














  • 6




    What's the order of $i$ then?
    – the_fox
    Jan 4 at 3:11






  • 2




    Does G have to have finite order?
    – Joel Pereira
    Jan 4 at 3:39










  • @JoelPereira:. isn't that implied by the statements $sum f(g) = n$ and $n = o(G)$?
    – Robert Lewis
    Jan 4 at 3:45










  • @the_fox :( back to square one.
    – Subhasis Biswas
    Jan 4 at 11:41






  • 1




    @RobertLewis no it's not implied. If G = the multiplicative group of $mathbb{R}^+$, we can still form the sum. In that case the sum would diverge.
    – Joel Pereira
    Jan 4 at 15:33








6




6




What's the order of $i$ then?
– the_fox
Jan 4 at 3:11




What's the order of $i$ then?
– the_fox
Jan 4 at 3:11




2




2




Does G have to have finite order?
– Joel Pereira
Jan 4 at 3:39




Does G have to have finite order?
– Joel Pereira
Jan 4 at 3:39












@JoelPereira:. isn't that implied by the statements $sum f(g) = n$ and $n = o(G)$?
– Robert Lewis
Jan 4 at 3:45




@JoelPereira:. isn't that implied by the statements $sum f(g) = n$ and $n = o(G)$?
– Robert Lewis
Jan 4 at 3:45












@the_fox :( back to square one.
– Subhasis Biswas
Jan 4 at 11:41




@the_fox :( back to square one.
– Subhasis Biswas
Jan 4 at 11:41




1




1




@RobertLewis no it's not implied. If G = the multiplicative group of $mathbb{R}^+$, we can still form the sum. In that case the sum would diverge.
– Joel Pereira
Jan 4 at 15:33




@RobertLewis no it's not implied. If G = the multiplicative group of $mathbb{R}^+$, we can still form the sum. In that case the sum would diverge.
– Joel Pereira
Jan 4 at 15:33










2 Answers
2






active

oldest

votes


















6














It is not necessary that $G$ be abelian, to wit:



If



$f(g) = 1, ; forall g in G, tag 1$



then clearly



$displaystyle sum_{g in G} f(g) = n, tag 2$



since



$o(G) = n; tag 3$



if



$exists h in G, ; f(h) ne 1, tag 4$



then since



$hG = G, tag 5$



we have



$$begin{align}
sum_{g in G} f(g) &= sum_{g in G} f(hg) \
&= sum_{g in G} f(h)f(g) \
&= f(h)sum_{g in G} f(g); tag 6
end{align}$$



with $f(h) ne 1$ this forces



$displaystyle sum_{g in G} f(g) = 0. tag 7$



$OEDelta$.






share|cite|improve this answer



















  • 1




    @Shaun: nice edit, thanks!
    – Robert Lewis
    Jan 4 at 6:23






  • 1




    Amazing. Just amazing.
    – Subhasis Biswas
    Jan 4 at 11:40










  • Interesting. I wonder if they just assumed abelian so that students could use the fundamental theorem. It's very easy to prove for cyclic groups.
    – Cameron Williams
    Jan 4 at 12:03










  • @SubhasisBiswas: thank you for your kind words. If you really like my answer, you might consider "accepting" it. Cheers!
    – Robert Lewis
    2 days ago



















0














Here is one novel way using representation theory. The homomorphism $f$ is a (1-dimensional) irreducible representation of a finite group $G$ and $sum_{g in G} f(g)$ is the sum of the character $chi_f$ over $g in G$, i.e. $chi_f = f$.



Since $sum_{g in G} chi_f(g) = lvert G rvertlangle chi_f, 1 rangle$, the sum is zero if and only if $1$ is not a direct summand of $f$. In that case, $f$ is trivial and the sum is $n$.






share|cite|improve this answer





















  • Can you please verify mine?
    – Subhasis Biswas
    Jan 4 at 6:00










  • the_fox in one of the earlier comments has already pointed out that a mistake in your proof is assuming the only elements with finite order in $mathbb{C}^*$ are $pm 1$, when in fact any complex $z$ such that $z^ell = 1$ for a nonzero integer $ell$, i.e. a root of unity, has finite order at most $ell$.
    – Riley
    Jan 4 at 7:01












  • $z^n=1$ does form a group. Now, can we somehow follow my approach to prove it?
    – Subhasis Biswas
    Jan 4 at 8:27










  • I'm not sure if I can adapt your approach. I might give this a go later myself, but you might be able to adapt it by first using the structure theorem for finitely generated abelian groups to first decompose $G$ into finite cyclic groups $mathbb{Z}_{m}$. On the direct summand $mathbb{Z}_{m}$, if $g_i$ is a generator, so that $g_i^m = 1$, then if we let $z = f(g_i)$ then $z^m = 1$ and $1 + z + cdots + z^{m-1} = 0$ if $z neq 1$.
    – Riley
    Jan 4 at 10:37










  • I was thinking exactly along this line. Now, conversion of this into isomorphism theorem would be really nice. :)
    – Subhasis Biswas
    Jan 4 at 11:40











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









6














It is not necessary that $G$ be abelian, to wit:



If



$f(g) = 1, ; forall g in G, tag 1$



then clearly



$displaystyle sum_{g in G} f(g) = n, tag 2$



since



$o(G) = n; tag 3$



if



$exists h in G, ; f(h) ne 1, tag 4$



then since



$hG = G, tag 5$



we have



$$begin{align}
sum_{g in G} f(g) &= sum_{g in G} f(hg) \
&= sum_{g in G} f(h)f(g) \
&= f(h)sum_{g in G} f(g); tag 6
end{align}$$



with $f(h) ne 1$ this forces



$displaystyle sum_{g in G} f(g) = 0. tag 7$



$OEDelta$.






share|cite|improve this answer



















  • 1




    @Shaun: nice edit, thanks!
    – Robert Lewis
    Jan 4 at 6:23






  • 1




    Amazing. Just amazing.
    – Subhasis Biswas
    Jan 4 at 11:40










  • Interesting. I wonder if they just assumed abelian so that students could use the fundamental theorem. It's very easy to prove for cyclic groups.
    – Cameron Williams
    Jan 4 at 12:03










  • @SubhasisBiswas: thank you for your kind words. If you really like my answer, you might consider "accepting" it. Cheers!
    – Robert Lewis
    2 days ago
















6














It is not necessary that $G$ be abelian, to wit:



If



$f(g) = 1, ; forall g in G, tag 1$



then clearly



$displaystyle sum_{g in G} f(g) = n, tag 2$



since



$o(G) = n; tag 3$



if



$exists h in G, ; f(h) ne 1, tag 4$



then since



$hG = G, tag 5$



we have



$$begin{align}
sum_{g in G} f(g) &= sum_{g in G} f(hg) \
&= sum_{g in G} f(h)f(g) \
&= f(h)sum_{g in G} f(g); tag 6
end{align}$$



with $f(h) ne 1$ this forces



$displaystyle sum_{g in G} f(g) = 0. tag 7$



$OEDelta$.






share|cite|improve this answer



















  • 1




    @Shaun: nice edit, thanks!
    – Robert Lewis
    Jan 4 at 6:23






  • 1




    Amazing. Just amazing.
    – Subhasis Biswas
    Jan 4 at 11:40










  • Interesting. I wonder if they just assumed abelian so that students could use the fundamental theorem. It's very easy to prove for cyclic groups.
    – Cameron Williams
    Jan 4 at 12:03










  • @SubhasisBiswas: thank you for your kind words. If you really like my answer, you might consider "accepting" it. Cheers!
    – Robert Lewis
    2 days ago














6












6








6






It is not necessary that $G$ be abelian, to wit:



If



$f(g) = 1, ; forall g in G, tag 1$



then clearly



$displaystyle sum_{g in G} f(g) = n, tag 2$



since



$o(G) = n; tag 3$



if



$exists h in G, ; f(h) ne 1, tag 4$



then since



$hG = G, tag 5$



we have



$$begin{align}
sum_{g in G} f(g) &= sum_{g in G} f(hg) \
&= sum_{g in G} f(h)f(g) \
&= f(h)sum_{g in G} f(g); tag 6
end{align}$$



with $f(h) ne 1$ this forces



$displaystyle sum_{g in G} f(g) = 0. tag 7$



$OEDelta$.






share|cite|improve this answer














It is not necessary that $G$ be abelian, to wit:



If



$f(g) = 1, ; forall g in G, tag 1$



then clearly



$displaystyle sum_{g in G} f(g) = n, tag 2$



since



$o(G) = n; tag 3$



if



$exists h in G, ; f(h) ne 1, tag 4$



then since



$hG = G, tag 5$



we have



$$begin{align}
sum_{g in G} f(g) &= sum_{g in G} f(hg) \
&= sum_{g in G} f(h)f(g) \
&= f(h)sum_{g in G} f(g); tag 6
end{align}$$



with $f(h) ne 1$ this forces



$displaystyle sum_{g in G} f(g) = 0. tag 7$



$OEDelta$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 4 at 6:13









Shaun

8,820113681




8,820113681










answered Jan 4 at 3:58









Robert LewisRobert Lewis

43.9k22963




43.9k22963








  • 1




    @Shaun: nice edit, thanks!
    – Robert Lewis
    Jan 4 at 6:23






  • 1




    Amazing. Just amazing.
    – Subhasis Biswas
    Jan 4 at 11:40










  • Interesting. I wonder if they just assumed abelian so that students could use the fundamental theorem. It's very easy to prove for cyclic groups.
    – Cameron Williams
    Jan 4 at 12:03










  • @SubhasisBiswas: thank you for your kind words. If you really like my answer, you might consider "accepting" it. Cheers!
    – Robert Lewis
    2 days ago














  • 1




    @Shaun: nice edit, thanks!
    – Robert Lewis
    Jan 4 at 6:23






  • 1




    Amazing. Just amazing.
    – Subhasis Biswas
    Jan 4 at 11:40










  • Interesting. I wonder if they just assumed abelian so that students could use the fundamental theorem. It's very easy to prove for cyclic groups.
    – Cameron Williams
    Jan 4 at 12:03










  • @SubhasisBiswas: thank you for your kind words. If you really like my answer, you might consider "accepting" it. Cheers!
    – Robert Lewis
    2 days ago








1




1




@Shaun: nice edit, thanks!
– Robert Lewis
Jan 4 at 6:23




@Shaun: nice edit, thanks!
– Robert Lewis
Jan 4 at 6:23




1




1




Amazing. Just amazing.
– Subhasis Biswas
Jan 4 at 11:40




Amazing. Just amazing.
– Subhasis Biswas
Jan 4 at 11:40












Interesting. I wonder if they just assumed abelian so that students could use the fundamental theorem. It's very easy to prove for cyclic groups.
– Cameron Williams
Jan 4 at 12:03




Interesting. I wonder if they just assumed abelian so that students could use the fundamental theorem. It's very easy to prove for cyclic groups.
– Cameron Williams
Jan 4 at 12:03












@SubhasisBiswas: thank you for your kind words. If you really like my answer, you might consider "accepting" it. Cheers!
– Robert Lewis
2 days ago




@SubhasisBiswas: thank you for your kind words. If you really like my answer, you might consider "accepting" it. Cheers!
– Robert Lewis
2 days ago











0














Here is one novel way using representation theory. The homomorphism $f$ is a (1-dimensional) irreducible representation of a finite group $G$ and $sum_{g in G} f(g)$ is the sum of the character $chi_f$ over $g in G$, i.e. $chi_f = f$.



Since $sum_{g in G} chi_f(g) = lvert G rvertlangle chi_f, 1 rangle$, the sum is zero if and only if $1$ is not a direct summand of $f$. In that case, $f$ is trivial and the sum is $n$.






share|cite|improve this answer





















  • Can you please verify mine?
    – Subhasis Biswas
    Jan 4 at 6:00










  • the_fox in one of the earlier comments has already pointed out that a mistake in your proof is assuming the only elements with finite order in $mathbb{C}^*$ are $pm 1$, when in fact any complex $z$ such that $z^ell = 1$ for a nonzero integer $ell$, i.e. a root of unity, has finite order at most $ell$.
    – Riley
    Jan 4 at 7:01












  • $z^n=1$ does form a group. Now, can we somehow follow my approach to prove it?
    – Subhasis Biswas
    Jan 4 at 8:27










  • I'm not sure if I can adapt your approach. I might give this a go later myself, but you might be able to adapt it by first using the structure theorem for finitely generated abelian groups to first decompose $G$ into finite cyclic groups $mathbb{Z}_{m}$. On the direct summand $mathbb{Z}_{m}$, if $g_i$ is a generator, so that $g_i^m = 1$, then if we let $z = f(g_i)$ then $z^m = 1$ and $1 + z + cdots + z^{m-1} = 0$ if $z neq 1$.
    – Riley
    Jan 4 at 10:37










  • I was thinking exactly along this line. Now, conversion of this into isomorphism theorem would be really nice. :)
    – Subhasis Biswas
    Jan 4 at 11:40
















0














Here is one novel way using representation theory. The homomorphism $f$ is a (1-dimensional) irreducible representation of a finite group $G$ and $sum_{g in G} f(g)$ is the sum of the character $chi_f$ over $g in G$, i.e. $chi_f = f$.



Since $sum_{g in G} chi_f(g) = lvert G rvertlangle chi_f, 1 rangle$, the sum is zero if and only if $1$ is not a direct summand of $f$. In that case, $f$ is trivial and the sum is $n$.






share|cite|improve this answer





















  • Can you please verify mine?
    – Subhasis Biswas
    Jan 4 at 6:00










  • the_fox in one of the earlier comments has already pointed out that a mistake in your proof is assuming the only elements with finite order in $mathbb{C}^*$ are $pm 1$, when in fact any complex $z$ such that $z^ell = 1$ for a nonzero integer $ell$, i.e. a root of unity, has finite order at most $ell$.
    – Riley
    Jan 4 at 7:01












  • $z^n=1$ does form a group. Now, can we somehow follow my approach to prove it?
    – Subhasis Biswas
    Jan 4 at 8:27










  • I'm not sure if I can adapt your approach. I might give this a go later myself, but you might be able to adapt it by first using the structure theorem for finitely generated abelian groups to first decompose $G$ into finite cyclic groups $mathbb{Z}_{m}$. On the direct summand $mathbb{Z}_{m}$, if $g_i$ is a generator, so that $g_i^m = 1$, then if we let $z = f(g_i)$ then $z^m = 1$ and $1 + z + cdots + z^{m-1} = 0$ if $z neq 1$.
    – Riley
    Jan 4 at 10:37










  • I was thinking exactly along this line. Now, conversion of this into isomorphism theorem would be really nice. :)
    – Subhasis Biswas
    Jan 4 at 11:40














0












0








0






Here is one novel way using representation theory. The homomorphism $f$ is a (1-dimensional) irreducible representation of a finite group $G$ and $sum_{g in G} f(g)$ is the sum of the character $chi_f$ over $g in G$, i.e. $chi_f = f$.



Since $sum_{g in G} chi_f(g) = lvert G rvertlangle chi_f, 1 rangle$, the sum is zero if and only if $1$ is not a direct summand of $f$. In that case, $f$ is trivial and the sum is $n$.






share|cite|improve this answer












Here is one novel way using representation theory. The homomorphism $f$ is a (1-dimensional) irreducible representation of a finite group $G$ and $sum_{g in G} f(g)$ is the sum of the character $chi_f$ over $g in G$, i.e. $chi_f = f$.



Since $sum_{g in G} chi_f(g) = lvert G rvertlangle chi_f, 1 rangle$, the sum is zero if and only if $1$ is not a direct summand of $f$. In that case, $f$ is trivial and the sum is $n$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 4 at 4:16









RileyRiley

1625




1625












  • Can you please verify mine?
    – Subhasis Biswas
    Jan 4 at 6:00










  • the_fox in one of the earlier comments has already pointed out that a mistake in your proof is assuming the only elements with finite order in $mathbb{C}^*$ are $pm 1$, when in fact any complex $z$ such that $z^ell = 1$ for a nonzero integer $ell$, i.e. a root of unity, has finite order at most $ell$.
    – Riley
    Jan 4 at 7:01












  • $z^n=1$ does form a group. Now, can we somehow follow my approach to prove it?
    – Subhasis Biswas
    Jan 4 at 8:27










  • I'm not sure if I can adapt your approach. I might give this a go later myself, but you might be able to adapt it by first using the structure theorem for finitely generated abelian groups to first decompose $G$ into finite cyclic groups $mathbb{Z}_{m}$. On the direct summand $mathbb{Z}_{m}$, if $g_i$ is a generator, so that $g_i^m = 1$, then if we let $z = f(g_i)$ then $z^m = 1$ and $1 + z + cdots + z^{m-1} = 0$ if $z neq 1$.
    – Riley
    Jan 4 at 10:37










  • I was thinking exactly along this line. Now, conversion of this into isomorphism theorem would be really nice. :)
    – Subhasis Biswas
    Jan 4 at 11:40


















  • Can you please verify mine?
    – Subhasis Biswas
    Jan 4 at 6:00










  • the_fox in one of the earlier comments has already pointed out that a mistake in your proof is assuming the only elements with finite order in $mathbb{C}^*$ are $pm 1$, when in fact any complex $z$ such that $z^ell = 1$ for a nonzero integer $ell$, i.e. a root of unity, has finite order at most $ell$.
    – Riley
    Jan 4 at 7:01












  • $z^n=1$ does form a group. Now, can we somehow follow my approach to prove it?
    – Subhasis Biswas
    Jan 4 at 8:27










  • I'm not sure if I can adapt your approach. I might give this a go later myself, but you might be able to adapt it by first using the structure theorem for finitely generated abelian groups to first decompose $G$ into finite cyclic groups $mathbb{Z}_{m}$. On the direct summand $mathbb{Z}_{m}$, if $g_i$ is a generator, so that $g_i^m = 1$, then if we let $z = f(g_i)$ then $z^m = 1$ and $1 + z + cdots + z^{m-1} = 0$ if $z neq 1$.
    – Riley
    Jan 4 at 10:37










  • I was thinking exactly along this line. Now, conversion of this into isomorphism theorem would be really nice. :)
    – Subhasis Biswas
    Jan 4 at 11:40
















Can you please verify mine?
– Subhasis Biswas
Jan 4 at 6:00




Can you please verify mine?
– Subhasis Biswas
Jan 4 at 6:00












the_fox in one of the earlier comments has already pointed out that a mistake in your proof is assuming the only elements with finite order in $mathbb{C}^*$ are $pm 1$, when in fact any complex $z$ such that $z^ell = 1$ for a nonzero integer $ell$, i.e. a root of unity, has finite order at most $ell$.
– Riley
Jan 4 at 7:01






the_fox in one of the earlier comments has already pointed out that a mistake in your proof is assuming the only elements with finite order in $mathbb{C}^*$ are $pm 1$, when in fact any complex $z$ such that $z^ell = 1$ for a nonzero integer $ell$, i.e. a root of unity, has finite order at most $ell$.
– Riley
Jan 4 at 7:01














$z^n=1$ does form a group. Now, can we somehow follow my approach to prove it?
– Subhasis Biswas
Jan 4 at 8:27




$z^n=1$ does form a group. Now, can we somehow follow my approach to prove it?
– Subhasis Biswas
Jan 4 at 8:27












I'm not sure if I can adapt your approach. I might give this a go later myself, but you might be able to adapt it by first using the structure theorem for finitely generated abelian groups to first decompose $G$ into finite cyclic groups $mathbb{Z}_{m}$. On the direct summand $mathbb{Z}_{m}$, if $g_i$ is a generator, so that $g_i^m = 1$, then if we let $z = f(g_i)$ then $z^m = 1$ and $1 + z + cdots + z^{m-1} = 0$ if $z neq 1$.
– Riley
Jan 4 at 10:37




I'm not sure if I can adapt your approach. I might give this a go later myself, but you might be able to adapt it by first using the structure theorem for finitely generated abelian groups to first decompose $G$ into finite cyclic groups $mathbb{Z}_{m}$. On the direct summand $mathbb{Z}_{m}$, if $g_i$ is a generator, so that $g_i^m = 1$, then if we let $z = f(g_i)$ then $z^m = 1$ and $1 + z + cdots + z^{m-1} = 0$ if $z neq 1$.
– Riley
Jan 4 at 10:37












I was thinking exactly along this line. Now, conversion of this into isomorphism theorem would be really nice. :)
– Subhasis Biswas
Jan 4 at 11:40




I was thinking exactly along this line. Now, conversion of this into isomorphism theorem would be really nice. :)
– Subhasis Biswas
Jan 4 at 11:40


















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