All distinct subgroups of $\mathbb{Z}_4 \times \mathbb{Z}_4$ isomorphic to $\mathbb{Z}_4$

This question is from a past exam.

Find all distinct subgroups of $\mathbb{Z}_4 \times \mathbb{Z}_4$ isomorphic to $\mathbb{Z}_4$

Attempt/Thoughts?

Since $\mathbb{Z}_4$ is cyclic we are looking for cyclic subgroups of the given group. Then can I use the fundamental theorem of f.g.ab. groups to solve this problem?. I have a hard time imagining how the elements in $\mathbb{Z}_4\times \mathbb{Z}_4$ look like. Is there a standard way to proceed in this type of problem?.

Can somebody help?. I usually don't ask for detailed answers in here. But in this case I would really appreciate it. Thanks.

The elements of the product are all pairs $(a,b)$ with $a$ and $b$ in $\mathbb{Z}_4$. Go patiently over all the candidates for generator of the subgroup isomorphic to $\mathbb{Z}_4$.

Start with $(0,1)$. It generates a subgroup of order $4$, consisting of $(0,1)$, $(0,2)$, $(0,3)$, $(0,0)$.

Note that $(0,2)$ is no good, while $(0,3)$ generates a group isomorphic to $\mathbb{Z}_4$ that we have already.

Now continue with $(1,1)$: good. Now try $(1,2)$: good, we get in turn $(2,0)$, $(3,2)$, $(0,0)$.

Continue. Analysis will quickly speed up. Be careful not to list the same subgroup twice. And take advantage of symmetry: the analysis of the group generated by $(0,a)$ is automatic once you have dealt with the group generated by $(a,0)$.

Remark: There is nothing wrong with computing. One becomes intimately acquainted with the structures that way. After a while we find shortcuts.

• This is exactly what I needed. Get my hands dirty and do these computations. Once I see the computations, I can actually understand what nrpeterson and julien are talking about. So Thank you. – minibuffer Jul 11 '13 at 18:56
• Nicholas: Just so you know I passed my exam :). Your answers/comments helped me a great deal. This one especially. "There is nothing wrong with computation" really struck a chord with me. MSE discourages commenting just to thank someone, but I couldn't resist. Thank you for sharing your knowledge. – minibuffer Sep 23 '13 at 6:28
• You are welcome, and congratulations. The abstractions can be tremendously helpful, but one needs to remember that they are ultimately about very concrete objects. – André Nicolas Sep 23 '13 at 6:34

HINT:

You're making this problem much too hard!

Elements of $\mathbb{Z}_4\times\mathbb{Z}_4$ are of the form $(a,b)$, where $a,b\in \mathbb{Z}_4$... and addition is defined coordinate-wise.

The real question here is: if you're handed $(a,b)$, where $a,b\in\mathbb{Z}_4$, what is its order? Essentially, you need to find all elements of order 4, then partition them up in to cyclic subgroups (since no two distinct cyclic subgroups of the same order can intersect).

Now, it turns out that the order of $(a,b)$ is the least common multiple of the order of $a$ and the order of $b$. So, since elements of $\mathbb{Z}_4$ can only have order 1, 2, or 4, it must be the case that either $a$ or $b$ has order 4; and, the converse also holds.

See if you can finish it up from there. Find an element; generate its subgroup; find another element that you haven't seen so far; etc.

If you want to count them without computing any particular subgroup, you can do it as follows.

An element $(x,y)\in \mathbb{Z}_4\times \mathbb{Z}_4$ has order $4$ if and only if one of $x,y$ has order $4$. This yields $16-4=12$ order $4$ elements (counting those for which both $x,y$ have order $1$ or $2$ and subtracting them).

A cyclic group of order $4$ has two generators (each of order $4$ of course). And every order $4$ element is the generator a cyclic order $4$ subgroup.

Two cyclic subgroups of order $4$ are equal if and only if they share a generator.

So there is a natural partition of the set of order $4$ elements into pairs of generators of the same order $4$ cyclic subgroup.

So there are $\frac{12}{2}=6$ distinct cyclic order $4$ subgroups. You will find what they actually are below if you really need it.

If you want to be more explicit, note that the map (isomorphism $-Id$ of $\mathbb{Z}_4\times \mathbb{Z}_4$, actually) $(x,y)\longmapsto (4-x,4-y)$ induces an involution of the set of order $4$ elements onto itself which sends the generator of a given cyclic order $4$ group to the other generator. This will help you go over all these $6$ subgroups injectively.

Details: $(0,1)$ and $(0,3)$ will be the generators of the first cyclic subgroup of order $4$ in lexicographic order. Then $(1,0)$ and $(3,0)$ . Then $(1,1)$ and $(3,3)$. Then $(1,2)$ and $(3,2)$. Then $(1,3)$ and $(3,1)$. Then $(2,1)$ and $(2,3)$. Then...there is nothing left.