For a subgroup $H$ of a finite group $G$ , when does $lvert operatorname{Aut}(H)rvert$ divide $lvert...












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Let $H$ be a subgroup of a finite group $G$. Is it true that $lvert operatorname{Aut}(H)rvert$ divides $lvert operatorname{Aut}(G)rvert$? What if we also assume $G$ is abelian? (I know that $lvert operatorname{Aut}(H)rvert space big| space lvert operatorname{Aut}(G)rvert$ if $G$ is cyclic).






group-theory finite-groups abelian-groups







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  • This is relevant: mathoverflow.net/questions/9749/…
    – hjhjhj57
    Apr 2 '15 at 5:31


















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Let $H$ be a subgroup of a finite group $G$. Is it true that $lvert operatorname{Aut}(H)rvert$ divides $lvert operatorname{Aut}(G)rvert$? What if we also assume $G$ is abelian? (I know that $lvert operatorname{Aut}(H)rvert space big| space lvert operatorname{Aut}(G)rvert$ if $G$ is cyclic).






group-theory finite-groups abelian-groups







share|cite|improve this question
























  • This is relevant: mathoverflow.net/questions/9749/…
    – hjhjhj57
    Apr 2 '15 at 5:31
















4












4








4


1





Let $H$ be a subgroup of a finite group $G$. Is it true that $lvert operatorname{Aut}(H)rvert$ divides $lvert operatorname{Aut}(G)rvert$? What if we also assume $G$ is abelian? (I know that $lvert operatorname{Aut}(H)rvert space big| space lvert operatorname{Aut}(G)rvert$ if $G$ is cyclic).






group-theory finite-groups abelian-groups







share|cite|improve this question















Let $H$ be a subgroup of a finite group $G$. Is it true that $lvert operatorname{Aut}(H)rvert$ divides $lvert operatorname{Aut}(G)rvert$? What if we also assume $G$ is abelian? (I know that $lvert operatorname{Aut}(H)rvert space big| space lvert operatorname{Aut}(G)rvert$ if $G$ is cyclic).






group-theory finite-groups abelian-groups




group-theory finite-groups abelian-groups






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share|cite|improve this question













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edited Jan 4 at 3:03









the_fox

2,48011431




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asked Apr 2 '15 at 4:50







user228168



















  • This is relevant: mathoverflow.net/questions/9749/…
    – hjhjhj57
    Apr 2 '15 at 5:31




















  • This is relevant: mathoverflow.net/questions/9749/…
    – hjhjhj57
    Apr 2 '15 at 5:31


















This is relevant: mathoverflow.net/questions/9749/…
– hjhjhj57
Apr 2 '15 at 5:31






This is relevant: mathoverflow.net/questions/9749/…
– hjhjhj57
Apr 2 '15 at 5:31












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It's not even true for abelian groups in general. Take $H=C_2times C_2$ as a subgroup of $G=C_4times C_2$. Then $lvert operatorname{Aut}(G)rvert=8$, while $lvert operatorname{Aut}(H)rvert=6$.






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    It's not even true for abelian groups in general. Take $H=C_2times C_2$ as a subgroup of $G=C_4times C_2$. Then $lvert operatorname{Aut}(G)rvert=8$, while $lvert operatorname{Aut}(H)rvert=6$.






    share|cite|improve this answer




























      6














      It's not even true for abelian groups in general. Take $H=C_2times C_2$ as a subgroup of $G=C_4times C_2$. Then $lvert operatorname{Aut}(G)rvert=8$, while $lvert operatorname{Aut}(H)rvert=6$.






      share|cite|improve this answer


























        6












        6








        6






        It's not even true for abelian groups in general. Take $H=C_2times C_2$ as a subgroup of $G=C_4times C_2$. Then $lvert operatorname{Aut}(G)rvert=8$, while $lvert operatorname{Aut}(H)rvert=6$.






        share|cite|improve this answer














        It's not even true for abelian groups in general. Take $H=C_2times C_2$ as a subgroup of $G=C_4times C_2$. Then $lvert operatorname{Aut}(G)rvert=8$, while $lvert operatorname{Aut}(H)rvert=6$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 4 at 3:06









        the_fox

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        answered Apr 2 '15 at 5:18









        verretverret

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        2,9841818






























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