Hi,
a) You need to observe that H is a subset of N by closure in H.
b) Looks okay.
c) In any finite group G, the order of any subgroup k divides the order of G. Apply this to H, N and G.
I would like to check if my proof of this proposition has been correctly done. I would also like help on proving part (c). Thanks in advance.
**Proposition:** Let $H$ be a subgroup of a group $G$, and let $N$ be the normalizer of $H$. Prove that:
(a) $H$ is a normal subgroup of $N$
(b) $H$ is a normal subgroup of $G$ if and only if $N$ = $G$
(c) $|H|$ divides $|N|$ and $|N|$ divides $|G|$.
**Proof:**
Part (a):
Suppose $H$ is a subgroup of $G$. Suppose $N$ is a normalizer of $H$, meaning $N(H)$ $=$ $\{g \in G: gHg^{-1}=H\}$. We want to show that $H$ is a normal subgroup of $N$. Note that by definition: $g$ $\in$ $N$ $\leftrightarrow gHg^{-1}=H$. Thus $gHg^{-1}=H$ for every $g$ $\in$ $N$. Hence, $H$ is normal in $N$.
Part (b): Suppose $H$ is a normal subgroup of $G$. Thus $gHg^{-1}=H$ for every $g$ $\in$ $G$. Thus for every $g$ in $G$, we have $gHg^{-1} = H$ $\in$ $N$. So $G$ is a subset of $N$. Since $N$ is a subset of $G$, we must have $N=G$.
Now other direction: Suppose $N=G$. Then, for every $g \in G$ and $g \in N$, we have: $gHg^{-1} = H$. Thus, by definition of normal, $H$ is normal in $G$.
How do I proceed with part c?
Thanks.