# Homework Help: Groups, order G = 60, G simple

1. May 7, 2007

### rourky

1. G a Group of order 60, G simple, prove G isomorphic to A5

2. Familiar with Sylow's Theorems, theorems leading up to Sylow.

3. We make the assumption that G is not isomorphic to A5
Then "given G cannot have a subgroup of index 2, 3, 4, 5," I can
get the result.
My problem is I don't know why the quoted statement is true.
Clear for 2 alright, given G simple, but 3, 4, 5?
Is this a fairly obvious result (if so, a hint please), or is it difficult to
prove and should i move on until I know more about groups?

2. May 7, 2007

### matt grime

Some of them are clear: if there is a subgroup of index d, then there is a homomorphism to S_d, the symmetric group on d elements - i.e. the (transitive, i.e. there is only one orbit) group action on the cosets. So for d=2,3,4, the orders of S_d are 2,6,24, so it is not injective, hence has a kernel, which would be normal. This only leaves the index 5 case. Again, this must be injective, so it is up to you to figure out what the (transitive) subgroups of order 60 are in S_5. There of course might be a better way to do the question than this.

3. May 7, 2007

### rourky

Thanks Matt,

A little "over my head". Not your fault though, just haven't heard of the terms "transitive subgroup" and "transitive group action". Unfamiliar with opening result as well, actually thought it was Cayley's theorem at first. Really appreciate the help, now happy in the knowledge that I wasn't overlooking some result from my course. Will leave topic until I know more about groups.

Thanks again, Ciaran

4. May 7, 2007

### matt grime

Erm. But the proof for 2,3,4 is elementary - it just needs you to know what a group homomorphism is. I'm sure you're happy with groupt actions on sets.

And as I said, there is no reason why what I wrote for d=5, or any other case, is the only way to do it. If your course set this question, you should be able to do it.

(Transitivity is a red herring, I just put it in there for completeness.)