Let $p$ be a prime number and let $G$ be a finite $p\text{-group}$. Let $M$ be a maximal subgroup of $G$.

Question

QUESTION: Let $p$ be a prime number and let $G$ be a finite $p\text{-group}$. Let $M$ be a maximal subgroup of $G$. Show that $M$ is a normal subgroup of $G$ and that $| G: M | = p$.

THE HINT GIVEN IS: By strong induction on $n$, where $| G | = p ^{n}$. Let $ y \in Z(G) - \{1 \}$, a convenient $x$ power belonging to $Z(G) - \{1\}$ has order $p$. Consider $G / \langle x \rangle$.

ANSWER GIVEN:

By induction on $n$, where $| G | = p^ n$. According to the tip, consider $x \in Z (G)$ of order $ p $ and let $N = \langle x \rangle$. The group $ G / N $ has order $ p^{n-1} $, so we can apply induction. If $ N $ is a subgroup of $ M $ then $ M / N $ is normal for the $ p $ index in $ G $. Now suppose $ N $ is not a subgroup of $ M $. Being $ M $ maximal we get $ NM = G $. On the other hand being $ | N | = p $ prime we have $ N \cap M = \{1 \} $ logo $ p^{n-1} = | G / N | = | MN / N | = | M / M \cap N | = | M | $ e we deduct $| G: M | = p $. In addition $ M $ is a subgroup of $ N_G (M) $ and $ N $ is a subgroup of $ N_G (M) $ because $ N \leq Z (G) $. It follows that $ G = NM \leq N_G (M) $ so $ M $ is a normal subset of $ G $.

MY QUESTIONS: I didn't understand the following steps showed in this proof.

1. The induction used in its solution.
2. If $N$ is subgroup of $M$ then one is stated that $M/N$ is normal (why?)
3. If $N$ is NOT a subgroup of $M$ then one is stated that $NM=G$ (again, why?)
4. Why $N \cap M=\{1\}?$
5. Why $N \leq N_G(M)$ ?
6. Why $NM\leq N_G(M)$?

Answer 1

1. The induction is on the exponent of the group. Every $p$ group has order $p^n$ for some $n$, so the induction is assuming that the theorem is true for every $p^m$ for $m<n$.

2. $N$ is part of the center so it commutes with everything, which is stronger than being normal.

3. If $N$ is not a subgroup of $M$ then some element of $NM$ must not be in $M$, which means $M$ is a subgroup of $NM$. $M$ is maximal, so that means $NM=G$.

4. $N$ is prime and cyclic so if any element of $N$ other than the identity is in $M$, then all of $N$ must be in $M$, which would make it a subgroup of $M$, which is assumed not to be true in this case.

5. Again, $N$ is part of the center, so it commutes with everything.

6. Of course $M$ normalizes itself, and $N$ commutes with everything in $M$ so it normalizes $M$. Therefore the product of these two groups normalizes $M$.