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. 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.
 A: 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.
A: Here is a proof.
Step  1. Prove it for the direct product $C_p\times C_p$ (find its subgroups of order $p$)
Step 2. Prove that if $f$ is a surjective homomorphism $G_1\to G_2$ then the number of maximal subgroups of $G_2$ is not bigger than the number of maximal subgroups of $G_1$ (the preimage of a maximal subgroup is a maximal subgroup).
Step 3. If $G$ is a non-trivial $p$-group then its center is non-trivial and contains a subgroup $C$ of order $p$. $C$ is normal in $G$.
Step 4.If $G/C$ is not cyclic then it has at least $p+1$ maximal subgroups. By 2. The same is true for $G$.
Step 5. If $G/C$ is cyclic then $G$ is Abelian.
Step 6. By the description of finite Abelian groups $G$ is a direct product of cyclic groups of orders $p^{k_1},...,p^{k_s}$ where $s>1$. Then $G$ has a homomorphism onto $C_p\times C_p$ and one can use 1 and 2.
