$G$ a non-cyclic group with $|G|=p^n$. Show that $G$ has at least $p + 3$ different subgroups.

Question

Question: Let $p$ be a prime integer, $n \in \Bbb{N}$, $n \gt 1$, and $G$ a non-cyclic group with $|G|=p^n$. Show that $G$ has at least $p + 3$ different subgroups.

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I know that since $G$ is non-cyclic, then there is no such $a \in$ $G$ for which $a^n=e$, $e$ being the neutral element of the group $G$. So every other element must belong to a subgroup that is not $G$.

Using Cauchy's Theorem, it means that we have an element $x \in$ $G$ for which we have $ord(x) = p$. And I assume that the assumption that must be made is that we have maximum $p+2$ subgroups.

But I do not really know how to go on from here? Could you give me some hints, or anything?

Answer 1

For $p\gt 2$, since $G$ is not cyclic, it contains a subgroup isomorphic to $C_p\times C_p$. Such a group has $p+1$ subgroups of order $p$, one of order $1$, and one of order $p^2$.

For $p=2$, the same holds unless $G$ is the quaternion group of order $8$, in which case you can verify directly it has at least 5 subgroups.

Answer 2

We will use the facts that $G/Z(G)$ is cyclic iff $G$ is abelian and that the center of a $p$-group is non-trivial.

Let's induct on the power of $p$ in the order of $G$. If the cardinality of $G$ is $p^2$ then since it's not cyclic it's the vector space of dimension two over $\mathbb{F}_p$. There are $p+1$ lines in this vector space, and also the two trivial subspaces, for a total of $p+3$.

Now suppose the theorem proven for groups of order $p^m$, $m < n$. Write $Z$ for the center of $G$. If $G$ is not abelian, then $G/Z$ is not cyclic and has strictly smaller cardinality than $G$, so it has at least $p+3$ subgroups and thus there are at least $p+3$ subgroups of $G$ containing $Z$ (and lots more too).

So we are reduced to the case that $G$ is abelian. Now since $G$ is not cyclic, it decomposes into a non-trivial product $G = H \times K$. If either $H$ or $K$ is not cyclic, we are done. Otherwise, there is a natural map from $H$ and from $K$ to the cyclic group of order $p$ (since both have subgroups of index $p$) and we have already verified the result for this space.