# 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)##.

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)##?

Last edited:
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