Intereting Posts

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proving that $\frac{(n^2)!}{(n!)^n}$ is an integer
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Combinatorial Proof
Every normal subgroup of a finite group is contained in some composition series
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I find this explanation in Isaacs’ *Algebra*:

Lemma.Let $H\subseteq G$ be a subgroup. Then $H$ is a normal subgroup if $H^g\subseteq H$ for all $g\in G$.The reader should be warned that this lemma does not state that $H^g=H$ whenever $H^g\subseteq H$. Since the inner automorphism induced by the element $g$ is a bijection, it is certainly true that $|H^g|=|H|$, and if $H$ is finite, this equality of orders together with the containment $H^g\subseteq H$ certainly does imply that $H^g=H$. For infinite subgroups, however, this does not follow and is not generally true.

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How does the lemma not state that? If $H^g\subseteq H$, then the lemma says that $H$ is a normal subgroup, which by definition means $H^g=H$. What am I missing?

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If $H^g\subseteq H$, then the lemma says that $H\trianglelefteq G$, which by definition means $H^g=H$.

With $g$ fixed, the condition $H^g\subseteq H$ does not imply $H^g=H$. It is true that *if* $H$ is normal, *then* the condition $H^g\subseteq H$ *does* imply $H^g=H$ (again, $g$ is some fixed element here), but it is arguable if it “follows from definition,” specifically the definition the text gives, because the text’s definition does not explicitly say anything about equality. Here is the proof:

- Suppose $H\trianglelefteq G$ is a normal subgroup and $H^g\subseteq H$ for some $g\in G$. Since $H$ is normal, we have $H^{g^{-1}}\subseteq H$; conjugate both sides by $g$ to obtain $H\subseteq H^g$. Since $H^g\subseteq H$ and $H\subseteq H^g$, we have obtained equality $H^g=H$.

Also note that, again with $g$ fixed, the lemma does *not* say that $H^g\subseteq H$ implies $H\trianglelefteq G$. Rather, the lemma says that if $H^g\subseteq H$ holds true *for all* $g\in G$, *then* $H$ is normal. The inclusion $H^g\subseteq H$ holding for some isolated individual element $g\in G$ is not enough for $H$ to be normal.

But here is what the text is getting at. In *general* (without a priori knowledge about normality), the condition $H^g\subseteq H$ does not imply $H^g=H$. That is, there exist situations where you have a subgroup $H\le G$ that *strictly* contains one of its conjugates $H^g$, i.e. $H^g\subset H$. In this case the $H$ is not normal, since $H^g\subset H\implies H\subset H^{g^{-1}}\implies H^{g^{-1}}\not\subseteq H$.

Let’s look at an example. Qiaochu gives a matrix example. Here is a semidirect product example. Take $\bf Q$ and $\bf Z$ to be additive groups, and define ${\bf Z}\to{\rm Aut}({\bf Q})$ so that $1\in{\bf Z}$ acts as e.g. the map $x\mapsto 2x$ on $\bf Q$. Then in $G={\bf Q}\rtimes{\bf Z}$, the subgroup $H=\langle1\rangle\subset \bf Q\hookrightarrow G$ conjugated by $1\in\bf Z$ is properly contained in $H$ – specifically, it is $2{\bf Z}\subsetneq H$.

Let $G$ be the group $\text{GL}_2(\mathbb{Q})$. This group has a subgroup $H \cong \mathbb{Z}$ consisting of all elements of the form $\left[ \begin{array}{cc} 1 & n \\ 0 & 1 \end{array} \right]$ where $n \in \mathbb{Z}$. Set $g = \left[ \begin{array}{cc} 2 & 0 \\ 0 & 1 \end{array} \right]$. Then $H^g = 2H$ is strictly contained in $H$.

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