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# Subgroups watch

1. Subgroups of d4: why are {i,r2,e1,e3} and {i,r2,e2,e4} where e is reflexion and r is rotation?

Any help would be appreciated.
2. A subgroup H is a subset of a group G that is itself a group under the same binary operation of G. So H is a subgroup if it satisfies the group axioms. Or, equivalently, H is a subgroup if
3. Just check that these sets are closed under composition. This is all you need since the group is finite.
4. (Original post by mathprof)
Just check that these sets are closed under composition. This is all you need since the group is finite.
I think you're right, but the "standard" method would be to show and I'd be nervous about claiming closure is enough without proving it.

[Sketch proof for the OP: Assume H closed and finite. Take a in H. Since H is finite, we can find m > n > 0 with a^m = a^n. Then a^(m-n) = e, so a^(m-n-1) = a^(-1), so the inverse of a is in H].
5. Have you done the cayley(multiplication) table? as others have suggested, draw it out, do the compositions and should be clear why there are subgroups!
6. I dont understand this either?

when you draw out a cayley table for the subgroups {I, R2, E1, E3} and {I, R2, E1, E4}
I get

I R2 E1 E3
R2 I E2 E4
E1 E2 I R1
E2 E1 R2 I , for the subgroup {I, R2, E1, E3} and

I R2 E2 E4
R2 I E1 E3
E2 E1 I R3
E4 E3 R3 I , for the subgroup {I, R2, E2, E4}

Please correct me if I'm wrong, but doesn't this suggest that there is no closure so therefore this cant possibly be subgroups?
7. (Original post by DFranklin)
I think you're right, but the &quot;standard&quot; method would be to show and I'd be nervous about claiming closure is enough without proving it.

[Sketch proof for the OP: Assume H closed and finite. Take a in H. Since H is finite, we can find m &gt; n &gt; 0 with a^m = a^n. Then a^(m-n) = e, so a^(m-n-1) = a^(-1), so the inverse of a is in H].
You're right that the method you've given is standard. In my OP, I forgot to add the two other conditions concerning the identity and inverses. After searching online to see what was necessary, I discovered that neither were when the group is finite. Your proof looks OK.
8. (Original post by belle_2106)
I dont understand this either?

when you draw out a cayley table for the subgroups {I, R2, E1, E3} and {I, R2, E1, E4}
I get

I R2 E1 E3
R2 I E2 E4
E1 E2 I R1
E2 E1 R2 I , for the subgroup {I, R2, E1, E3} and

I R2 E2 E4
R2 I E1 E3
E2 E1 I R3
E4 E3 R3 I , for the subgroup {I, R2, E2, E4}

Please correct me if I'm wrong, but doesn't this suggest that there is no closure so therefore this cant possibly be subgroups?
What is meant by e1, e2, e3, etc here? These are relections about what axes? I was assuming that in the OP they were powers of a single reflection e.
9. (Original post by mathprof)
What is meant by e1, e2, e3, etc here? These are relections about what axes? I was assuming that in the OP they were powers of a single reflection e.
That's a good point I didn't think about E1, E2 etc being any of the reflections,
in my example I used E1 as the y=-x reflection, E2 as y=x, E3 along the y axis and E4 along the x axis, hence why I was confused to how the subgroup listed in the original question could be a subgroup.
I didn't take into account in each reflection you could use any notation.
and yes they are all reflections of e.
10. Now that I think about it, the original poster can't mean powers of a reflection because for any reflection. Furthermore, I don't think it matters what the specific reflections are because the result is true regardless. In your case, however, I'll need to know what your notation stands for in order to find the error in the Cayley table. We are talking about the symmetry group of a square and your square is positioned in the first quadrant with a vertex at the origin and the sides lying on the axes, correct?

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Updated: April 8, 2011
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