Question: Let $G$ be a finite $p$-group, where $p$ is prime. Show that if $G$ is not cyclic, then $G$ has at least $p+1$ maximal subgroups.
I've seen this thread: Let $G$ be a finite $p$-group with has more than one maximal subgroup. Prove that $G$ has at least $p+1$ maximal subgroups. but I wasn't sure that if the condition "more than one maximal subgroup" could replace "$G$ is not cyclic". So….
Let $|G|=p^n$, for prime $p$ and $n\in\mathbb{N}$. Since $G$ is not cyclic, there does not exist $g\in G$ such that $g^n=1$. Therefore, every element belongs to a maximal subgroup, so $G=\cup M$, where $M$ is a maximal subgroup (just taking the union of all maximal subgroups). Now, $|M|=p^{n-1}$, since $G$ is a $p$-group. Suppose there exist at most $p$ maximal subgroups. So, I now want to add up the orders of all the maximal subgroups and show that it is less than $|G|$. That is, we have $p^{n-1}+p^{n-1}-1+\dots+p^{n-1}-p$… but I am not getting the contradiction that I want. Can anyone help me fix this? Thank you.
Best Answer
If you have only $p$ subgroups of order at most $p^{n-1}$, because each of them has the group identity as a common element, their union can only have size at most $1+p(p^{n-1}-1) = p^n-p+1 \lt p^n$. But you have already shown that every element of $G$ must be in some maximal subgroup, so the union must have size $p^n$. That's a contradiction, so there must be at least $p+1$ maximal subgroups.