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Until recently, I believed that a subgroup of a direct product was the direct product of subgroups. Obviously, there exists a trivial counterexample to this statement.

I have a question regarding this. It is threefold:

  1. I hear that a theorem called Goursat's lemma states characterization of subgroups of a direct product. I read it on Lang's Algebra, but its statement is a bit counterintuitive and I do not know how to use it to solve problems. I would be grateful if you could give a hint regarding this lemma.

  2. I had the aforementioned false belief when I tried to solve this problem, and it seemed to work in this specific problem. When is the generally false statement hold? (Or is my reasoning in concluding the answer to the problem is 8 flawed?)

  3. I have problems similar to the previous one, namely, counting the subgroups of $\Bbb Z/p\Bbb Z \times \Bbb Z/p \Bbb Z$ and $\Bbb Z/p^2\Bbb Z \times \Bbb Z/p^2 \Bbb Z$, where $p$ is a prime number. How can I use the previous considerations to solve this problem?

I would appreciate your help.

EDIT: I fixed a typo in the last problem.

EDIT 2: The problem I mentioned in 2. is actually about the number of elements of some order, not that of subgroups. (I forgot about that when I wrote this question.) I am still interested in the question stated in 2., though.

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    $\begingroup$ I don't have time to give a full answer right now. But from the characterization, one can see that all subgroups being such direct products is equivalent to the two groups not having a non-trivial common subquotient. $\endgroup$ – Tobias Kildetoft Sep 6 '13 at 8:19
  • $\begingroup$ it's a nontrivial question for intuitive purposes. best way i've found to think about it is through the automorphism group. $\endgroup$ – Alexander Gruber Sep 6 '13 at 8:41
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    $\begingroup$ @AlexanderGruber Could you please elaborate on that? $\endgroup$ – Pteromys Sep 7 '13 at 11:36
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I will first state the characterization of the subgroup of a direct product. Then I will try to make this characterization seem at least a bit intuitive by explaining what sorts of subgroups it gives. Finally, I will demonstrate how to use it for finding the number of subgroups of $\mathbb{Z}/p^2\mathbb{Z}\times\mathbb{Z}/p^2\mathbb{Z}$.

The characterization says the following: Let $G$ and $H$ be groups. Then the subgroups of $G\times H$ are in bijection with the set of $5$-tuples $(G_1,G_2,H_1,H_2,\varphi)$ where $G_2\unlhd G_1\leq G$, $H_2\unlhd H_1\leq H$ and $\varphi: G_1/G_2\to H_1/H_2$ is an isomorphism.
The correspondence is given by assigning to such a tuple the subgroup $\{(g,h)\in G_1\times H_1\mid \varphi(gG_2) = hH_2\}$.

So let us take a look at what sort of subgroups we can construct "naively" in the direct product, and try to put them on the form above.
First, we have those of the form $G_1\times H_1$ (these are the easiest one to think of). These correspond to picking $G_2 = G_1$ and $H_2 = H_1$ and the unique homomorphism from $G_1/G_2$ to $H_1/H_2$ (since these groups are both trivial). (Check that this does indeed produce the subgroup $G_1\times H_1$).

Next, the common example of a subgroup not of this form is in the direct product $G\times G$ where we have the diagonal subgroup given by all elements of the form $(g,g)$. To make this fit in better with the above, assume that $G$ and $H$ are isomorphic and that $\varphi$ is an isomorphism between them (writing it as $G\times G$ then essentially corresponds to having picked an isomorphism). Now, the diagonal above corresponds to the subgroup consisting of all elements of the form $(g,\varphi(g))$, and we see that this subgroup corresponds to picking $G_1 = G$, $H_1 = H$, $G_2 = \{e\}$, $H_2 = \{e\}$ and $\varphi$ as the isomorphism (again, check this).

But we can of course easily generalize the above if the two groups have subgroups $G_1$ and $H_1$ which are isomorphic via some isomorphism $\varphi$, giving us the subgroup consisting of elements of the form $(g,\varphi(g))$, now with $g\in G_1$ and $h\in H_1$. This corresponds to picking $G_1$ and $H_1$ to be those subgroups (hence the choice of notation), and $G_2$ and $H_2$ both to be the trivial subgroup like above (and again the isomorphism should be $\varphi$).

Generalizing further on the above, let us assume that we have two subgroups $G_1$ and $H_1$ and a surjective homomorphism $\varphi':G_1\to H_1$. Now we can again form the subgroup consisting of all elements of the form $(g,\varphi'(g))$ for $g\in G_1$ (we never needed the homomorphism to be bijective above). This corresponds to picking $G_2 = \rm{ker}(\varphi')$, $H_2 = \{e\}$ and $\varphi$ to be the isomorphism $G_1/G_2\to H_1$ induced by $\varphi'$ (check this).

Note that the reason we pick $\varphi'$ to be surjective in the above is that otherwise, we might as well just look at the image of $\varphi'$.

The final step in generalizing these ideas is probably the least intuitive. Namely, we have two subgroups $G_1$ and $H_1$, but we do not necessarily have any homomorphism between them. Instead, we have a homomorphism $\varphi$ from $G_1$ to some quotient $H_1/H_2$ which is surjective (again, the surjectivity is just because otherwise, we would restrict to the image). Let us first assume this map is also injective, to make notations a bit simpler. In this case, we wish to form something like the subgroups from above, but we can no longer just take something like $(g,\varphi(g))$, since $\varphi(g)$ is no longer an element of $H_1$ but an element of $H_1/H_2$. On the other hand, $\varphi(g)$ is a coset of $H_2$ in $H_1$ and is therefore a subset of $H_1$, so for each $g\in G$ we can take all the elements of the form $(g,h)$ where $h$ is in the coset $\varphi(g)$ of $H_2$ in $H_1$. One should again check that this is indeed a subgroup, and that this corresponds to picking $G_1 = \{e\}$ and and the rest are already in the appropriate notation.

To get rid of the requirement of injectivity above, we set $G_2$ to be the kernel of the homomorphism and replace the homomorphism by the induced isomorphism $G_1/G_2\to H_1/H_2$. The pairs we now need are those of the form $(g,h)$ where again $h$ is in the coset $\varphi(g)$ but now we need to interpret $\varphi(g)$ as $\varphi(gG_2)$ (since $\varphi$ need not even be defined on $G_1$). So our elements are those of the form $(g,h)$ where $\varphi(gG_2) = hH_2$. This is fortunately precisely the form given by the characterization, so we now have all the possible subgroups.

Let's now, finally, take a look at finding the number of subgroups of $\mathbb{Z}/p^2\mathbb{Z}\times \mathbb{Z}/p^2\mathbb{Z}$.

So we need to find all possible $5$-tuples as above. Since each of the factors have precisely $3$ subgroups, all isomorphism classes of subgroups and quotients are determined by their order, and the tuples need the two quotients to be isomorphic, we get a total of $14$ types of tuples ($9$ where the order is $1$, $4$ where the order is $p$ and $1$ where the order is $p^2$), and for each we then need to know how many isomorphisms we can pick.

To make the notation a bit easier, denote the group by $G\times H$. Let $G'$ and $H'$ be the proper, non-trivial subgroups of $G$ and $H$ respectively. Denote by $1$ the trivial subgroup of either.

Our tuples then have the forms given below. The number after each tuple is the number of possible isomorphisms $\varphi$ we can pick (this number is the order of the automorphism group of $G_1/G_2$ which is of course also the order of the automorphism group of $H_1/H_2$. In our cases, these automorphism groups are easy to calculate as the possible quotients are all cyclic).
$(G,G,H,H,\varphi)$ $1$
$(G,G,H',H',\varphi)$ $1$
$(G,G,1,1,\varphi)$ $1$
$(G,G',H,H',\varphi)$ $p-1$
$(G,G',H',1,\varphi)$ $p-1$
$(G,1,H,1,\varphi)$ $p^2 - p$
$(G',G',H,H,\varphi)$ $1$
$(G',G',H',H',\varphi)$ $1$
$(G',G',1,1,\varphi)$ $1$
$(G',1,H,H',\varphi)$ $p-1$
$(G',1,H',1,\varphi)$ $p-1$
$(1,1,H,H,\varphi)$ $1$
$(1,1,H',H',\varphi)$ $1$
$(1,1,1,1,\varphi)$ $1$

Now it is just a matter of adding these number together to see that the total number of subgroups is $(p^2 - p) + 4(p-1) + 9 = p^2 + 3p + 5$.

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  • $\begingroup$ Thank you. Is the characterization you described the same thing as what is called Goursat's lemma? $\endgroup$ – Pteromys Sep 11 '13 at 0:11
  • $\begingroup$ @Pteromys Yes (at least, I know it was proven by Goursat, so I suppose it must be the one). $\endgroup$ – Tobias Kildetoft Sep 11 '13 at 8:06
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I want to answer question three, as per your above comment. I do not use Goursat's lemma explicitly. Sorry.

Before starting, I should say that I use the phrase "by symmetry" a few times. This basically means "If $G=H\times K$ then I have proven it for $H$ so it must hold for $K$".

The first thing to notice is the following lemma:

Lemma: Suppose $G=H\times K$. A subgroup $S\leq G$ has the form $\{(h_i, k_i): i\in I\}$ where the elements $\{h_i\}_I$ form a subgroup of $H$ and the elements $\{k_i\}$ form a subgroup of $K$.

Proof: To see this, recall that $H$ is normal in $G$ and the quotient group is simply $K$, so the image of $S$ contains precisely the elements $\{k_i\}$, as required. By symmetry, we have proven the lemma.

Question 3)i) I will write $\mathbb{Z}_p$ for $\mathbb{Z}/p\mathbb{Z}$. So, the first part of question three asks "What are the subgroups of $G=\mathbb{Z}_p\times\mathbb{Z}_p$?" I'll write $G=H\times K$. There are precisely $(p-1)+2=p+1$ proper, non-trivial subgroups of $G$. To see this, note that $\mathbb{Z}_p$ has no proper, non-trivial subgroups, and so writing $S=\{(h_i, k_i): i\in I\}$ we have one of three situations occurring: either $h_i$ is trivial and $k_i$ is non-trivial for all $i\in I$, or $h_i$ is non-trivial and $k_i$ is trivial for all $i\in I$, or both $h_i$ and $k_i$ are non-trivial for all $I$. In the first two situations the subgroups are isomorphic to $\mathbb{Z}_p$ and are non-equal (why?). The last situation is most interesting. Here, note that an arbitrary, non-trivial element $h_i$ generates $H$, and so the element $(h_i, k_i)$ generates a subgroup for all $k_i\in K$. There are $p-1$ such non-trivial $k_i\in K$, so there are $p-1$ such subgroups. Note that these subgroups are all non-equal. Therefore, there are $2+(p-1)=p+1$ proper, non-trivial subgroups of $G$. Note that if $p=2$ then $p+1=3$, which tallys with the fact that the Klein 4-group has three proper, non-trivial subgroups.

Question 3)ii) For $G=\mathbb{Z}_{p^2}\times\mathbb{Z}_p^2=H\times K$, use the same trick as above. However, $\mathbb{Z}_{p^2}$ has a proper, non-trivial subgroup, namely $\mathbb{Z}_{p^2}$, and so when we pick the elements $\{h_i\}$ there are three choices, as they are either generators for $H$ ($p$ choices), generators for the proper subgroup ($p-1$ choices), or are trivial ($1$ choice). Moreover, here there is a proper, non-trivial subgroup which is a direct product!

We shall classify the cyclic subgroups (up to symmetry). Note that there is only one proper, non-trivial subgroup which is not cyclic. The "up to symmetry" means that we have classified them but that you cannot simply add up the numbers and multiply by two. You will have counted stuff twice if you do this.

Case 1: Suppose $h_i$ is trivial. Then there are two non-trivial subgroups associated to this case, corresponding to the non-trivial subgroups of $K$.

Case 2: Suppose $h_i$ has order $p$. Then $h_i$ can be paired with every element $k_i$ of order $p^2$ in $K$ to form a subgroup of order $p^2$. Note that these are maximal but that they are not all non-equal. There are $p-1=p(p-1)/p$ such subgroups (why?). On the other hand, $h_i$ can be paired with every element $k_i$ of order $p$ in $K$ to form a subgroup of order $p$. There are $p-1$ such subgroups. Finally, $h_i$ can be paired with the identity to form a subgroup of order $p$ in $K$. There is one such subgroup.

Case 3: Suppose $h_i$ has order $p^2$. Then $h_i$ can be paired with every element $k_i$ of $K$ to form a subgroup of order $p^2$. Note that these are maximal but are again not all non-equal. There are $p^2+(p-1)+p=p^2+2p-1$ such subgroups (why?).

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  • $\begingroup$ Thank you. I would be most grateful if you could make the argument more formal, especially its latter half. I am not sure what you are doing there. $\endgroup$ – Pteromys Sep 9 '13 at 6:02
  • $\begingroup$ @Pteromys I prefer giving not-quite-complete answers here, because I feel that it helps people understand them better (it also saves me time, which is why I am side-stepping your request just now...). If you have any specific questions though I would be more than happy to answer them. $\endgroup$ – user1729 Sep 10 '13 at 9:35

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