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

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 $$G / \langle x \rangle$$.

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 $$NM\leq N_G(M)$$?

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.

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 $$M$$.

• Thank you very much for explain for me. I just still not undestanding item number 4. Apr 21, 2020 at 11:26