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Finding Subgroups



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

{H,}\{H, *\} is a group since its Cayley table shows us that the group is closed under the operation *, specifically, it shows us that a,bH,(ab)H\forall a,b \in H, (a * b) \in H. We can also see the existence of the identity element ee. 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:

a12=a2a_1^2 = a_2

a13=a3a_1^3 = a_3

a14=ea_1^4 = e

Then {e,a1,a2,a3}\{e, a_1, a_2, a_3 \} forms a subgroup of order 4.

The only other element in H that is of order 4 is a3a_3 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! :smile:
(edited 9 years ago)
Original post by Zacken


The only other element in H that is of order 4 is a3a_3 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! :smile:


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.
Reply 2
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 Z2×Z2\mathbb{Z}_2 \times \mathbb{Z}_2?

Thing is, I have no idea what Z2×Z2\mathbb{Z}_2 \times \mathbb{Z}_2 is. :frown:

Okay, so the elements of Z2×Z2\mathbb{Z}_2 \times \mathbb{Z}_2 are (0,0),(1,0),(1,1)(0,0), (1,0), (1,1).

I can't see anyway to link this to my subgroup.
(edited 9 years ago)
Original post by Zacken
So, there's a subgroup that is isomorphic to the group Z2×Z2\mathbb{Z}_2 \times \mathbb{Z}_2?

Thing is, I have no idea what Z2×Z2\mathbb{Z}_2 \times \mathbb{Z}_2 is. :frown:


This

Have a look at the b elements in your given group as a starting point.
Reply 4
Original post by ghostwalker
This

Have a look at the b elements in your given group as a starting point.


I think that e,a2,b1,b2{e, a_2, b_1, b_2} 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? :smile:
Original post by Zacken
I think that e,a2,b1,b2{e, a_2, b_1, b_2} 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? :smile:


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.
(edited 9 years ago)
Reply 6
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 a2a_2 come in, is that because it has order 2 as well?
Original post by Zacken
Okay, I can go with that, but where does the a2a_2 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 b1b2=a2b_1b_2=a_2
Reply 8
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 b1b2=a2b_1b_2=a_2


I see, thanks for the help! PRSOM. :smile:

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