# If H is a subgroup, then H is subgroup of normalizer

## Homework Statement

Show that if $H$ is a subgroup of $G$, then $H \le N_G (H)$

## The Attempt at a Solution

Essentially, we need to show that $H \subseteq N_G (H)$; since they are both groups under the same binary operation the fact that they are subgroups will result. So let $h \in H$. We want to show that $h \in N_G (H)$, i.e. we want to show that $hHh^{-1} = H$. But since the map $\sigma_h (x) = hxh^{-1}$ is automorphism an and thus a permutation from $H$ to $H$, the set $hHh^{-1}$ is just a permuted version of $H$, and so by set theory $hHh^{-1} = H$, and $h \in N_G (H)$.

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fresh_42
Mentor
Isn't the point to show that $H \trianglelefteq N_G(H)$ or even better, that $N_G(H)$ is the largest subgroup with this property? For $H \leq N_G(H)$ you only need to write $hHh^{-1} \subseteq H$ which is trivial for (sub-)groups, no conjugation or automorphism needed, just the group operation.

• Mr Davis 97
Isn't the point to show that $H \trianglelefteq N_G(H)$ or even better, that $N_G(H)$ is the largest subgroup with this property? For $H \leq N_G(H)$ you only need to write $hHh^{-1} \subseteq H$ which is trivial for (sub-)groups, no conjugation or automorphism needed, just the group operation.
I'm just reading the problem word for word. I haven't learned about normal subgroups yet, so maybe that's why is seems trivial. But, one question. Why does $hHh^{-1} \subseteq H$ imply that $H \le N_G (H)$?

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fresh_42
Mentor
$N_G(H)=\{\,g\in G\,:\,gHg^{-1}\subseteq H\,\}$ per definition and this implies automatically that it contains $H$ because $H$ is a group.

A normal subgroup $N$ is one which has the property that $gNg^{-1} \subseteq N$ for all $g\in G\,.$ The clue is that normal subgroups are exactly those for which there can be defined a group structure on the set of equivalence classes $\{\,[g]\,:\,[g]=gN\; , \;g \in G\,\}$. With an ordinary subgroup this cannot be done, with a normal subgroup it can (well definition is what fails with ordinary subgroups). The definition of $N_G(H)$ also shows that $H$ is normal in $N_G(H)$ - per construction, and this is why it is called normaizer of $H$ in $G$. It's also the largest such subgroup for otherwise an element $gNg^{-1}$ would already be in $N_G(H)$.

• Mr Davis 97
$N_G(H)=\{\,g\in G\,:\,gHg^{-1}\subseteq H\,\}$ per definition and this implies automatically that it contains $H$ because $H$ is a group.

A normal subgroup $N$ is one which has the property that $gNg^{-1} \subseteq N$ for all $g\in G\,.$ The clue is that normal subgroups are exactly those for which there can be defined a group structure on the set of equivalence classes $\{\,[g]\,:\,[g]=gN\; , \;g \in G\,\}$. With an ordinary subgroup this cannot be done, with a normal subgroup it can (well definition is what fails with ordinary subgroups). The definition of $N_G(H)$ also shows that $H$ is normal in $N_G(H)$ - per construction, and this is why it is called normaizer of $H$ in $G$. It's also the largest such subgroup for otherwise an element $gNg^{-1}$ would already be in $N_G(H)$.
Ah I see. Also, one thing, Dummit and Foote defines $N_G (A) = \{ g \in G ~|~ gAg^{-1} = A \}$. Not sure why it is equality and not subset.

fresh_42
Mentor
Ah I see. Also, one thing, Dummit and Foote defines $N_G (A) = \{ g \in G ~|~ gAg^{-1} = A \}$. Not sure why it is equality and not subset.
This doesn't make a difference: $gNg^{-1} \subseteq N \Longrightarrow N \subseteq g^{-1}Ng \subseteq N$ as the relation is true for $g$ as well as $g^{-1}$.

• Mr Davis 97