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# Finding Subgroups Watch

1. I've been trying this question, pretty sure I've gotten (i) right, if not, could you just take a look?

is a group since its Cayley table shows us that the group is closed under the operation , specifically, it shows us that . We can also see the existence of the identity element . It is also apparent that every element has an inverse since the identity element appears in every row/column of the Cayley table.

It's with (ii) that I have a major problem. I found one subgroup through the usual method, that is:

Then forms a subgroup of order 4.

The only other element in H that is of order 4 is which generates the same subgroup. So, I'm confused as to how to find the next subgroup. Does my method of finding a subgroup only work for cyclic subgroups? I'd be grateful for any clearing up on this, thanks!
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2. (Original post by Zacken)

The only other element in H that is of order 4 is which generates the same subgroup. So, I'm confused as to how to find the next subgroup. Does my method of finding a subgroup only work for cycling subgroups? I'd be grateful for any clearing up on this, thanks!
Yep, that will only work for cyclic subgroups.

If you recall, there are two groups of order 4 (up to isomorphism), one is the cyclic group Z_4, and the other is the group Z_2 X Z_2 - which has no element of order 4.

I'd start with one of the b elements and combine it with something else.
3. (Original post by ghostwalker)
Yep, that will only work for cyclic subgroups.

If you recall, there are two groups of order 4 (up to isomorphism), one is the cyclic group Z_4, and the other is the group Z_2 X Z_2 - which has no element of order 4.

I'd start with one of the b elements and combine it with something else.
So, there's a subgroup that is isomorphic to the group ?

Thing is, I have no idea what is.

Okay, so the elements of are .

I can't see anyway to link this to my subgroup.
4. (Original post by Zacken)
So, there's a subgroup that is isomorphic to the group ?

Thing is, I have no idea what is.
This

Have a look at the b elements in your given group as a starting point.
5. (Original post by ghostwalker)
This

Have a look at the b elements in your given group as a starting point.
I think that is a subgroup, but I stumbled upon that on dumb luck, I'm not really understanding what you're trying to put across. Mind going through your thinking process step by step?
6. (Original post by Zacken)
I think that is a subgroup, but I stumbled upon that on dumb luck, I'm not really understanding what you're trying to put across. Mind going through your thinking process step by step?
Say we choose b_1. Giving us the subgroup e, b_1. So 2 elements, but we need 4. So, lets add to it. Choose b_2, another element of order 2, and we end up with the subgroup you have.

My method is more inspired guesswork, rather than anything formal. But note the two elements of order 2, as per the Z_2 X Z_2 group.
7. (Original post by ghostwalker)
Say we choose b_1. Giving us the subgroup e, b_1. So 2 elements, but we need 4. So, lets add to it. Choose b_2, another element of order 2, and we end up with the subgroup you have.

My method is more inspired guesswork, rather than anything formal. But note the two elements of order 2, as per the Z_2 X Z_2 group.
Okay, I can go with that, but where does the come in, is that because it has order 2 as well?
8. (Original post by Zacken)
Okay, I can go with that, but where does the come in, is that because it has order 2 as well?
Having added b_2 to your set, then you need to add all the combinations to make it a subgroup, and
9. (Original post by ghostwalker)
Having added b_2 to your set, then you need to add all the combinations to make it a subgroup, and
I see, thanks for the help! PRSOM.

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