# Finding the subgroups of direct product group

1. Jul 9, 2017

### TimeRip496

1. The problem statement, all variables and given/known data
What are the subgroups of Z2 x Z2 x Z2?

2. Relevant equations
Hint: There are 16 subgroups.

3. The attempt at a solution
So far I only manage to get 15 and I am not even sure if these are correct.
My answer: $$(0,0,0) , (Z_2,Z_2,Z_2), (1,1,1), (0,0,1), (0,1,0), (1,0,0), (0,1,1), (1,0,1), (1,1,0), (0,Z_2,Z_2), (Z_2,0,Z_2),(Z_2,Z_2,0), (1,1,Z_2), (1,Z_2,1), (Z_2,1,1)$$

where <(Z2, Z2, Z2)> is not the same as <(1,1,1)>
$$<(Z_2,Z_2,Z_2)> = {(0,0,0),(1,1,1),(0,0,1),(0,1,0),(1,0,0),(0,1,1),(1,0,1),(1,1,0)}$$
$$<(1,1,1)>={(0,0,0),(1,1,1)}$$

Forgive me as I just started group theory and I am now using videos on visual group theory as a guide. The above question comes from here.

2. Jul 9, 2017

### Dick

Your notation is not at all clear. Since the group has order 8 a subgroup will have order 1, 2, 4 or 8. Why don't you try and count them by order? Order 1 and 8 should be easy. Order 2 subgroups are pretty easy to describe. Order 4 takes a little more work (remember there are only two group structures of order 4 and since there are no elements of order 4 the subgroup must look like the Klein group {0, x, y, x+y}).

3. Jul 9, 2017

### Staff: Mentor

Your notation is a bit unusual, but this is mainly because we are trained to consider subgroups up to group bijections and your "diagonal" groups like $(1,1,1)\, , \,(1,1,0)\, , \,(1,1,\mathbb{Z}_2)$ as only images which are "the same" as $\mathbb{Z}_2,\mathbb{Z}_2,\mathbb{Z}_2^2$ and would prefer to denote the embeddings less short.

I couldn't find another group either and also got only 15 groups. So we've both missed the same subgroup or the number 16 is wrong. Why do you think it are 16?

4. Jul 9, 2017

### Dick

I have one subgroup of order 1, one of order 8, seven of order 2 and seven of order 4. Totals to 16.

5. Jul 9, 2017

### Staff: Mentor

I have only six of order four. I can't see a symmetry broken with those, as it is the case with the all-3-diagonal at order two. But don't tell yet, I want to search for the lost candidate

Edit: Got it. To write down the four elements instead of the scheme used in the OP made the difference.

Last edited: Jul 9, 2017
6. Jul 10, 2017

### TimeRip496

Sorry for the scheme I used cause I just started group theory and I am following what I see from the video on Visual Group Theory above. Still I don't know what the last subgroup is so I am hoping you will enlighten me here. Thanks

7. Jul 10, 2017

### Dick

I'm guessing it's {(0,0,0), (1,1,0), (1,0,1), (0,1,1)}.

8. Jul 10, 2017

### Staff: Mentor

No need for a sorry. It works better than I thought. However, it also led to overlook the missing group (see @Dick's post above), because it isn't covered by the scheme. One way to get help on these kind of questions are the following tables I frequently use when in doubt:
https://en.wikipedia.org/wiki/List_of_small_groups
https://de.wikipedia.org/wiki/Liste_kleiner_Gruppen

The second one is in the wrong language, but as a list of groups, this is a minor disadvantage. It lists the groups a bit differently, so I use both of them depending on the individual case and the mood I'm in.