Is a Subgroup Characteristic in its Normalizer?

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Let $G$ be a finite group and $H subseteq G$. Is it true that $H$ is a characteristic subgroup of $N_G(H)$? Knowing that "the something" subgroup must be characteristic, I believe it must be true.
Any comments would be appreciated!






abstract-algebra group-theory finite-groups







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  • Note that $H$ is not "the something" subgroup of $N_G(H)$ here...
    – verret
    3 hours ago














up vote
3
down vote

favorite












Let $G$ be a finite group and $H subseteq G$. Is it true that $H$ is a characteristic subgroup of $N_G(H)$? Knowing that "the something" subgroup must be characteristic, I believe it must be true.
Any comments would be appreciated!






abstract-algebra group-theory finite-groups







share|cite|improve this question





















  • Note that $H$ is not "the something" subgroup of $N_G(H)$ here...
    – verret
    3 hours ago












up vote
3
down vote

favorite









up vote
3
down vote

favorite











Let $G$ be a finite group and $H subseteq G$. Is it true that $H$ is a characteristic subgroup of $N_G(H)$? Knowing that "the something" subgroup must be characteristic, I believe it must be true.
Any comments would be appreciated!






abstract-algebra group-theory finite-groups







share|cite|improve this question













Let $G$ be a finite group and $H subseteq G$. Is it true that $H$ is a characteristic subgroup of $N_G(H)$? Knowing that "the something" subgroup must be characteristic, I believe it must be true.
Any comments would be appreciated!






abstract-algebra group-theory finite-groups




abstract-algebra group-theory finite-groups






share|cite|improve this question













share|cite|improve this question











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asked 5 hours ago









Sean

48129




48129











  • Note that $H$ is not "the something" subgroup of $N_G(H)$ here...
    – verret
    3 hours ago
















  • Note that $H$ is not "the something" subgroup of $N_G(H)$ here...
    – verret
    3 hours ago















Note that $H$ is not "the something" subgroup of $N_G(H)$ here...
– verret
3 hours ago




Note that $H$ is not "the something" subgroup of $N_G(H)$ here...
– verret
3 hours ago










2 Answers
2






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3
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Let $G=C_ptimes C_p$ and $H$ any subgroup of order $p$. Then $N_G(H)=G$
but $H$ is not characteristic in $G$.






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  • Nice example! thank you1
    – Sean
    5 hours ago

















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3
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The normalizer of any subgroup of an abelian group is the whole group, but there are subgroups of abelian groups that are not characteristic.






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  • Thank you for the counterexample!
    – Sean
    5 hours ago










  • @Sean No problem.
    – Matt Samuel
    5 hours ago










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2 Answers
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active

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2 Answers
2






active

oldest

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active

oldest

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active

oldest

votes








up vote
3
down vote













Let $G=C_ptimes C_p$ and $H$ any subgroup of order $p$. Then $N_G(H)=G$
but $H$ is not characteristic in $G$.






share|cite|improve this answer




















  • Nice example! thank you1
    – Sean
    5 hours ago














up vote
3
down vote













Let $G=C_ptimes C_p$ and $H$ any subgroup of order $p$. Then $N_G(H)=G$
but $H$ is not characteristic in $G$.






share|cite|improve this answer




















  • Nice example! thank you1
    – Sean
    5 hours ago












up vote
3
down vote










up vote
3
down vote









Let $G=C_ptimes C_p$ and $H$ any subgroup of order $p$. Then $N_G(H)=G$
but $H$ is not characteristic in $G$.






share|cite|improve this answer












Let $G=C_ptimes C_p$ and $H$ any subgroup of order $p$. Then $N_G(H)=G$
but $H$ is not characteristic in $G$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 5 hours ago









Lord Shark the Unknown

94.5k956123




94.5k956123











  • Nice example! thank you1
    – Sean
    5 hours ago
















  • Nice example! thank you1
    – Sean
    5 hours ago















Nice example! thank you1
– Sean
5 hours ago




Nice example! thank you1
– Sean
5 hours ago










up vote
3
down vote













The normalizer of any subgroup of an abelian group is the whole group, but there are subgroups of abelian groups that are not characteristic.






share|cite|improve this answer




















  • Thank you for the counterexample!
    – Sean
    5 hours ago










  • @Sean No problem.
    – Matt Samuel
    5 hours ago














up vote
3
down vote













The normalizer of any subgroup of an abelian group is the whole group, but there are subgroups of abelian groups that are not characteristic.






share|cite|improve this answer




















  • Thank you for the counterexample!
    – Sean
    5 hours ago










  • @Sean No problem.
    – Matt Samuel
    5 hours ago












up vote
3
down vote










up vote
3
down vote









The normalizer of any subgroup of an abelian group is the whole group, but there are subgroups of abelian groups that are not characteristic.






share|cite|improve this answer












The normalizer of any subgroup of an abelian group is the whole group, but there are subgroups of abelian groups that are not characteristic.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 5 hours ago









Matt Samuel

35.8k63461




35.8k63461











  • Thank you for the counterexample!
    – Sean
    5 hours ago










  • @Sean No problem.
    – Matt Samuel
    5 hours ago
















  • Thank you for the counterexample!
    – Sean
    5 hours ago










  • @Sean No problem.
    – Matt Samuel
    5 hours ago















Thank you for the counterexample!
– Sean
5 hours ago




Thank you for the counterexample!
– Sean
5 hours ago












@Sean No problem.
– Matt Samuel
5 hours ago




@Sean No problem.
– Matt Samuel
5 hours ago

















 

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