# How to prove that two groups have the same lattice of subgroups?

If p is an odd prime , and $P1 = Z_{p} \times Z_{p^2}$ and P2 is the non-abelian group of order $p^3$ with an element of order $p^2$, how would you go about proving that the two groups have the same lattice of subgroups? Is it easy to describe all subgroups and hence show they are isomorphic? We are currently on the chapter dealing with upper and lower central series and solvable groups.

## 2 Answers

The two groups you mentioned above are clearly not isomorphic since the first is abelian while the second is not. The right way to think about it is to write down the corresponding subgroups indeed (for $P_1$ is easy), and exploit the fact that the second group $P_2$ can be described up to isomorphism as $\mathbb{Z}_{p^2} \rtimes \mathbb{Z}_{p}$. That could be a useful exercise to do so!

$\mathbf{EDIT 1:}$ I expand my answer a little bit more for pedagogical reasons mostly in order to underline the fact that we have something a bit impressive here (at least at a first glance for a beginner). What I have written above is that we have two $p$-groups, which are the same sets and following up your comment that you are doing in class upper-lower central series I say the following. If two $p$-groups have isomorphic lattices of subgroups then they necessarily have the same cardinality. Why is that? (try to prove it) Additionally they have the following pathology: Although they are not isomorphic, they have the same, in some sense combinatorial structure. So we have $p$-groups with equal order (in fact equal sets) with equal number of subgroups which are not isomorphic...

$\mathbf{EDIT 2:}$ After your recent comment, let me delineate one way to find/count subgroups. (Might be a bit scary in a first glance, buts it's worth-reading I guess).

$\mathbf{History, Source:}$ The problem of finding subgroups in a direct product of arbitrary groups $G_1 \times G_2$ has stated many years ago and has been solved/characterized by Goursat (commonly referred in bibliography as Gourat's Lemma). Wikipedia page might be useful and help you clarify some things, although I haven't came across this lemma in a common textbook so far and I know this result for a very specific reason. So if someone knows a textbook that treats this result might be useful for you to write out the title. (the proof might be non-trivial for a beginner, but you can think just the statement of it which is easy).

$\mathbf{Lemma:}$ Assume that $G_1$, $G_2$ are two groups, and let by $G$ denote the direct product of them, $G_1 \times G_2$. Then there is one-to-one correspondence between the subgroups of $G$, and 5-tuples of the form $(K_1, K_2, L_1, L_2, \sigma)$, where $K_2\unlhd K_1\leq G_1$, $L_2\unlhd L_1\leq G_2$, and $\sigma : K_1/K_2 \rightarrow L_1/L_2$ is an isomorphism between them.

So in your case we have $G_1 = \mathbb{Z}_p$, $G_2 = \mathbb{Z}_{p^2}$, hence you have to understand how these 5-tuples look like in that case. However, for general groups or complicated ones the above lemma in practise might be difficult to be applied, though in your case the groups are not only cyclic, but even better they are $p$-groups, therefore the number subgroups is a bit limited and their nature well understood. Try to work out along with Goursat's Lemma and if you can't I will provide some additional idea.

$\mathbf{P.S.}$ I haven't checked online, but likely you can find your answer somewhere if you check around for those two groups explicitly I believe, since this is a relatively easy case where we can count the subgroups directly.

• What would be all the subgroups of P1, and how in general do you find all subgroups a direct product? Is there a technique to use? Commented Oct 24, 2017 at 3:58

$$P_{2}$$ has presentation \begin{align*} \left\langle y\right\rangle\rtimes\left\langle x\right\rangle &=\left\langle\left. x,y\right|\ x^{p}=y^{p^{2}}=1,\ xyx^{-1}=y^{p+1}\right\rangle \end{align*} and the kernel of the $$p$$-th power map is the abelian subgroup $$\left\langle y^{p}\right\rangle\times\left\langle x\right\rangle$$. Then $$\left(xy^{k}\right)^{p}$$ is a non-trivial element of $$Z\left(P\right)$$ (see Exercise~5.4.9) for $$1\leq k. That is, $$\left|\left\langle xy^{k}\right\rangle\right|>p$$ so $$\left|\left\langle xy^{k}\right\rangle\right|=p^{2}$$. Consider the maximal subgroups of $$P_{1}$$ and $$P_{2}$$. Of these, both $$P_{1}$$ and $$P_{2}$$ have one elementary abelian subgroup and $$p$$ cyclic subgroups. Then the sublattices of the maximal subgroups are identical so $$P_{1}$$ and $$P_{2}$$ have the same lattice of subgroups.