Prove $|H|$ Divides $|N|$ and $|N|$ Divides $|G|$: Proof Verification

[kalish1]

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.

 

[jvicens]

Your proof for parts (a) and (b) looks correct. For part (c), here is a possible approach:

To prove that $|H|$ divides $|N|$, we can consider the cosets of $H$ in $N$ and use the fact that $N(H)$ is a subgroup of $N$. Specifically, since $N(H)$ is a subgroup of $N$, we know that $|N(H)|$ divides $|N|$. But $N(H)$ is the set of all elements in $N$ that normalize $H$, so we can write $N(H)$ as a union of cosets of $H$ in $N$ (by definition of cosets). Therefore, $|N(H)|$ is equal to the number of cosets of $H$ in $N$, which is denoted by $[N]$. So we have $|N(H)| = [N] \cdot |H|$. Since $|N(H)|$ divides $|N|$, we have $[N] \cdot |H|$ divides $|N|$, which means $|H|$ divides $|N|$.

To prove that $|N|$ divides $|G|$, we can use a similar approach. We know that $N$ is a subgroup of $G$, so $|N|$ divides $|G|$ by Lagrange's theorem. But we also know that $N$ is the set of all elements in $G$ that normalize $H$, so we can write $N$ as a union of cosets of $H$ in $G$. Therefore, $|N|$ is equal to the number of cosets of $H$ in $G$, which is denoted by $[G]$. So we have $|N| = [G] \cdot |H|$. Since $|N|$ divides $|G|$ and $|H|$ divides $|N|$, we can conclude that $|N|$ divides $|G|$.

I hope this helps. Let me know if you have any further questions or if you would like me to clarify any part of the proof.