Here is the problem I want to solve:
Let $H$ be a subgroup of order $2$ in $G.$ Show that $N_{G}(H) = C_G(H).$
And here is my solution to it:
Remember that, the normalizer of $H$ in $G$ is defined as $N_G(H) = \{g \in G| gHg^{-1} = H\}$ where $gHg^{-1} = \{ghg^{-1} | h \in H\}.$ Also, the centralizer of $H$ in $G$ is defined as $C_G(H) = \{g \in G| ghg^{-1} = h \text{ for all } h \in H\}$\
Let $H$ be a subgroup of order $2$ in the group $G.$ We want to show that $N_{G}(H) = C_G(H).$
But clearly, $C_G(H) \subseteq N_{G}(H)$ by the definition of both. Now, it remains to show that $N_{G}(H) \subseteq C_G(H).$
Suppose $g \in N_G(H).$ Since $H$ is a subgroup of order $2$ in $G,$ then $H$ contains the identity element $e$ and only one other element $h \neq e$ i.e., $H = \{e, h\}.$ So, by the definition of the $N_{G}(H)$ and since $g \in N_G(H),$ we have $\{g e g^{-1} = e, g h g^{-1} = h\}$ i.e., $\{g e g^{-1}, g h g^{-1} \} = \{e, h\} = H$ which means that $g h g^{-1} = h$ for every $h \in H.$ Therefore, we get, by the definition of $C_G(H)$ that $g \in C_G(H)$. Hence $N_{G}(H) \subseteq C_G(H)$ as required.
I am not sure if the my justification in bold above is correct or no, could anyone check this for me please? as my justification made me feel that the definition of $C_G(H)$ and $N_G(H)$ are the same.
Best Answer
You are right. To see why they differ, assume there are three elements in $H$. Then $N_G(H)$ is all of the elements of $g$ that stabilize the set $H$ under conjugation. If $H=\{e,x,x^2\}$, then certainly $e$ is always centralized, not just normalized. But it could be that some element of $G$ conjugates $x$ to $x^2$ (and vice versa). That would normalize the subgroup $H$ but not centralize it.
This goes on explicitly in the symmetric group $S_3$. Here $H=\{1,(1,2,3),(1,3,2)\}$ is normalized by the whole group, but $(1,2)$ swaps by conjugation the two $3$-cycles.
But in the case $|H|=2$, $H=\{e,x\}$. Again, $e$ is always centralized by anything normalizing $H$, but now $x$ doesn't have anywhere to go in $H$ other than itself. So anything normalizing $H$ must centralize both elements of it, i.e., centralize $H$ as a whole.