Finding the subgroups of direct product group

[TimeRip496]

Homework Statement

What are the subgroups of Z₂ x Z₂ x Z₂?

Homework Equations

Hint: There are 16 subgroups.

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.

 

[Dick]

Homework Statement

What are the subgroups of Z₂ x Z₂ x Z₂?

Homework Equations

Hint: There are 16 subgroups.

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.

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}).

 

[fresh_42]

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?

 

[Dick]

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?

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

 

[fresh_42]

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.

 

[TimeRip496]

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.

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

 

[Dick]

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

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

 

[fresh_42]

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

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.