How can I prove $a$ had infinite order $Rightarrow$ $f(a)$ has infinite order? [duplicate]
This question already has an answer here:
Isomorphism $f$ preserves the order of an element?
3 answers
Problem: Let $f : G → H$ be an isomorphism. Prove that if $a in G$
has infinite order, then so does $f(a)$, and if $a$ has finite order
$n$, then so does $f(a)$. Conclude that if $G$ has an element of some
order $n$ and $H$ does not, then $G notcong H$.
My proof: If $a$ has finite order $n$, then $a^n=e Rightarrow f(a^n) = f(a)^n = e$. So, $f(a)$ has order $n$.
How can I prove $a$ had infinite order $Rightarrow$ $f(a)$ has infinite order?
abstract-algebra group-theory
asked Jan 4 at 14:05
MinhMinh
1788
marked as duplicate by Dietrich Burde abstract-algebra
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Jan 4 at 14:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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show 4 more comments
This question already has an answer here:
Isomorphism $f$ preserves the order of an element?
3 answers
Problem: Let $f : G → H$ be an isomorphism. Prove that if $a in G$
has infinite order, then so does $f(a)$, and if $a$ has finite order
$n$, then so does $f(a)$. Conclude that if $G$ has an element of some
order $n$ and $H$ does not, then $G notcong H$.
My proof: If $a$ has finite order $n$, then $a^n=e Rightarrow f(a^n) = f(a)^n = e$. So, $f(a)$ has order $n$.
How can I prove $a$ had infinite order $Rightarrow$ $f(a)$ has infinite order?
abstract-algebra group-theory
asked Jan 4 at 14:05
MinhMinh
1788
marked as duplicate by Dietrich Burde abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
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Jan 4 at 14:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
Have you considered the contrapositive statement?
– Shaun
Jan 4 at 14:06
I know the contrapositive statement, but I couldn't describe it on my proof.
– Minh
Jan 4 at 14:08
The contrapositive statement give $f(a).f(a) dots neq e$
– Minh
Jan 4 at 14:09
1
Since $f$ is an isomorphism, can't you use $f^{-1}$ combined with the thing you already showed? That should give you the contrapositive unless I glossed over something.
– tilper
Jan 4 at 14:10
1
@AshishK Are we need to prove $n$ is the least positive integer with $f(a)^n = e$?
– Minh
Jan 4 at 14:21
|
show 4 more comments
This question already has an answer here:
Isomorphism $f$ preserves the order of an element?
3 answers
Problem: Let $f : G → H$ be an isomorphism. Prove that if $a in G$
has infinite order, then so does $f(a)$, and if $a$ has finite order
$n$, then so does $f(a)$. Conclude that if $G$ has an element of some
order $n$ and $H$ does not, then $G notcong H$.
My proof: If $a$ has finite order $n$, then $a^n=e Rightarrow f(a^n) = f(a)^n = e$. So, $f(a)$ has order $n$.
How can I prove $a$ had infinite order $Rightarrow$ $f(a)$ has infinite order?
abstract-algebra group-theory
asked Jan 4 at 14:05
MinhMinh
1788
This question already has an answer here:
Isomorphism $f$ preserves the order of an element?
3 answers
Problem: Let $f : G → H$ be an isomorphism. Prove that if $a in G$
has infinite order, then so does $f(a)$, and if $a$ has finite order
$n$, then so does $f(a)$. Conclude that if $G$ has an element of some
order $n$ and $H$ does not, then $G notcong H$.
My proof: If $a$ has finite order $n$, then $a^n=e Rightarrow f(a^n) = f(a)^n = e$. So, $f(a)$ has order $n$.
How can I prove $a$ had infinite order $Rightarrow$ $f(a)$ has infinite order?
This question already has an answer here:
Isomorphism $f$ preserves the order of an element?
3 answers
abstract-algebra group-theory
abstract-algebra group-theory
asked Jan 4 at 14:05
MinhMinh
1788
asked Jan 4 at 14:05
MinhMinh
1788
asked Jan 4 at 14:05
MinhMinh
1788
asked Jan 4 at 14:05
MinhMinh
1788
asked Jan 4 at 14:05
MinhMinh
1788
1788
marked as duplicate by Dietrich Burde abstract-algebra
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Jan 4 at 14:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Dietrich Burde abstract-algebra
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Jan 4 at 14:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
Have you considered the contrapositive statement?
– Shaun
Jan 4 at 14:06
I know the contrapositive statement, but I couldn't describe it on my proof.
– Minh
Jan 4 at 14:08
The contrapositive statement give $f(a).f(a) dots neq e$
– Minh
Jan 4 at 14:09
1
Since $f$ is an isomorphism, can't you use $f^{-1}$ combined with the thing you already showed? That should give you the contrapositive unless I glossed over something.
– tilper
Jan 4 at 14:10
1
@AshishK Are we need to prove $n$ is the least positive integer with $f(a)^n = e$?
– Minh
Jan 4 at 14:21
|
show 4 more comments
1
Have you considered the contrapositive statement?
– Shaun
Jan 4 at 14:06
I know the contrapositive statement, but I couldn't describe it on my proof.
– Minh
Jan 4 at 14:08
The contrapositive statement give $f(a).f(a) dots neq e$
– Minh
Jan 4 at 14:09
1
Since $f$ is an isomorphism, can't you use $f^{-1}$ combined with the thing you already showed? That should give you the contrapositive unless I glossed over something.
– tilper
Jan 4 at 14:10
1
@AshishK Are we need to prove $n$ is the least positive integer with $f(a)^n = e$?
– Minh
Jan 4 at 14:21
1
1
Have you considered the contrapositive statement?
– Shaun
Jan 4 at 14:06
Have you considered the contrapositive statement?
– Shaun
Jan 4 at 14:06
I know the contrapositive statement, but I couldn't describe it on my proof.
– Minh
Jan 4 at 14:08
I know the contrapositive statement, but I couldn't describe it on my proof.
– Minh
Jan 4 at 14:08
The contrapositive statement give $f(a).f(a) dots neq e$
– Minh
Jan 4 at 14:09
The contrapositive statement give $f(a).f(a) dots neq e$
– Minh
Jan 4 at 14:09
1
1
Since $f$ is an isomorphism, can't you use $f^{-1}$ combined with the thing you already showed? That should give you the contrapositive unless I glossed over something.
– tilper
Jan 4 at 14:10
Since $f$ is an isomorphism, can't you use $f^{-1}$ combined with the thing you already showed? That should give you the contrapositive unless I glossed over something.
– tilper
Jan 4 at 14:10
1
1
@AshishK Are we need to prove $n$ is the least positive integer with $f(a)^n = e$?
– Minh
Jan 4 at 14:21
@AshishK Are we need to prove $n$ is the least positive integer with $f(a)^n = e$?
– Minh
Jan 4 at 14:21
|
show 4 more comments
1 Answer
1
active
oldest
votes
Your proof works in the other direction as well:
$$
e = f(a)^n = f(a^n) overset{f text{ isomorphism}}{implies} e = f^{-1}(e) = a^n
$$
answered Jan 4 at 14:12
bruderjakob17bruderjakob17
1537
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Your proof works in the other direction as well:
$$
e = f(a)^n = f(a^n) overset{f text{ isomorphism}}{implies} e = f^{-1}(e) = a^n
$$
answered Jan 4 at 14:12
bruderjakob17bruderjakob17
1537
add a comment |
Your proof works in the other direction as well:
$$
e = f(a)^n = f(a^n) overset{f text{ isomorphism}}{implies} e = f^{-1}(e) = a^n
$$
answered Jan 4 at 14:12
bruderjakob17bruderjakob17
1537
add a comment |
Your proof works in the other direction as well:
$$
e = f(a)^n = f(a^n) overset{f text{ isomorphism}}{implies} e = f^{-1}(e) = a^n
$$
answered Jan 4 at 14:12
bruderjakob17bruderjakob17
1537
Your proof works in the other direction as well:
$$
e = f(a)^n = f(a^n) overset{f text{ isomorphism}}{implies} e = f^{-1}(e) = a^n
$$
answered Jan 4 at 14:12
bruderjakob17bruderjakob17
1537
answered Jan 4 at 14:12
bruderjakob17bruderjakob17
1537
answered Jan 4 at 14:12
bruderjakob17bruderjakob17
1537
answered Jan 4 at 14:12
bruderjakob17bruderjakob17
1537
1537
add a comment |
add a comment |
1
Have you considered the contrapositive statement?
– Shaun
Jan 4 at 14:06
I know the contrapositive statement, but I couldn't describe it on my proof.
– Minh
Jan 4 at 14:08
The contrapositive statement give $f(a).f(a) dots neq e$
– Minh
Jan 4 at 14:09
1
Since $f$ is an isomorphism, can't you use $f^{-1}$ combined with the thing you already showed? That should give you the contrapositive unless I glossed over something.
– tilper
Jan 4 at 14:10
1
@AshishK Are we need to prove $n$ is the least positive integer with $f(a)^n = e$?
– Minh
Jan 4 at 14:21