1. ## Subroups...

I have this problem... Let H = {B in S7 | B(2)=2 and B(5)=5}. Prove that H is a subgroup.

I have tried to do this by saying that the remaining cycle is a 5 cycle which is even...and I think I am way off base.

Is there anyone that can help get me started? I guess I just have to show non-empty, AB is in there and that B^-1 is in there, but truly, I am struggling with the permutation thing.

Thanks

2. Originally Posted by ginafara
I have this problem... Let H = {B in S7 | B(2)=2 and B(5)=5}. Prove that H is a subgroup.

I have tried to do this by saying that the remaining cycle is a 5 cycle which is even...and I think I am way off base.

Is there anyone that can help get me started? I guess I just have to show non-empty, AB is in there and that B^-1 is in there, but truly, I am struggling with the permutation thing.

Thanks
I do not understand $H = \{ \beta \in S_7 | \beta(2) = 2 \mbox{ and }\beta (5) = 5 \}$. Do you mean that say that $\beta$ maps 2 onto 2 and 5 onto 5.

3. Yes, that is what it means. I copied the problem exactly as written in the text.

Sadly, I find regular groups much easier than the permutation groups.

4. So H is the set of permutations that fix 2 and 5.

H is closed in S7 (under the operation of composition) because if you take h1 and h2 in H, h1h2(2) = h1(2) = 2, and similarly, h1h2(5) = h1(5) = 5. So h1h2 is in H.
Obviously the identity element (the permutation that fixes everything) is in H, and this also shows it's nonempty.
And the inverse of every h in H is also in H, because if h1 fixes 2 and 5, then its inverse fixes them as well.. If not, then the h1 composed with its inverse wouldn't give the identity back.
That should be enough to prove it's a subgroup.

5. or you can also use the subgroup test, if you have already discussed (and proved it).. Ü