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A subgroup $H$ is normal if and only if $\forall a \in G (H = aHa^{-1})$, $a H a^{-1} \subseteq H$ and for arbitrary $a \in G$, $x \mapsto axa^{-1}$ is an isomorphism, why doesn't this imply that $H$ is normal in $G$?

Shaun
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victor hernandez
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    I'm not sure I understand the question, but it could be because "isomorphic" and "the same subset" are not the same. – Jason DeVito - on hiatus Apr 20 '21 at 18:51
  • For finite groups, conjugation by a fixed element is always an isomorphism, but there exist plenty of finite groups with non-normal subgroups. Your statement would imply that every subgroup of every finite group is normal, which isn't true at all. For the details, see: https://math.stackexchange.com/questions/250264/is-the-conjugation-map-always-an-isomorphism – morrowmh Apr 20 '21 at 18:52
  • I can't make any sense of this question at all. – Derek Holt Apr 20 '21 at 19:33

3 Answers3

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Because if $\varphi\colon G\longrightarrow G$ is an isomorphism, it is not necessarily true that, for each subgroup $H$ of $G$, $\varphi(H)\subset H$, not even if $\varphi(x)=axa^{-1}$ for some $a\in G$. For instance, if $G$ is the group of all $2\times2$ invertible real matrices and $H$ is the group of those matrices of $G$ of the form $\left[\begin{smallmatrix}a&b\\0&c\end{smallmatrix}\right]$, then$$\begin{bmatrix}0&1\\-1&0\end{bmatrix}.\begin{bmatrix}a&b\\0&c\end{bmatrix}.\begin{bmatrix}0&1\\-1&0\end{bmatrix}^{-1}=\begin{bmatrix}c&0\\-b&a\end{bmatrix}.$$

José Carlos Santos
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All this means is that the image of $H$ under that isomorphism is isomorphic to $H$. It need not be $H$ itself. For example, let $G = S_3$, $H = \{ e, (12)\}$, and $a = (123)$. Then $$a(12)a^{-1} = (123)(12)(132) = (23),$$ so that $aHa^{-1} = \{ e, (23)\}$.

rogerl
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That $x\mapsto axa^{-1} $ is an isomorphism is always true. It's a so-called _inner automorphism _.

This doesn't make any subgroup normal, we do have that $aHa^{-1}\cong H $ though.

Actually all you need is a non-normal subgroup of any group to see it.

What you actually need for normality is that $H $ is invariant under all of the inner automorphisms.