# Groups of order 60 and elements of order 5

1. Aug 10, 2007

### djxl

1. The problem statement, all variables and given/known data

Let G be a group with order $$\left| G \right| = 60$$. Assume that G is simple.

Now let H be the set of all elements that can be written as a product of elements of order 5 in G. Show that H is a normal subgroup of G. Then conclude that H = G

2. Relevant equations

3. The attempt at a solution

I started by proving that H acutally is a subgroup.

I've then shown that there are 6 Sylow-5 subgroups in G and that they are cyclic. I know that all the elements of order 5 are the generators of the Sylow-5 subgroups. But how I can use that to show that H is normal escapes me.

All help/hints appreciated :).

2. Aug 10, 2007

### matt grime

Sylow, won't help you, I don't think - the elements of order 5 do not generate Sylow-5 subgroups. The product of two elements of order 5 is not necessarily an element of order 5 (or any power of 5).

H is trivially a subgroup - there is nothing to prove there. What about normality? This is straight forward - conjugation preserves order, and notice that

gxyg^{-1} = gxg^{1-}gyg^{-1}

3. Aug 10, 2007

### learningphysics

Count the total number of elements in all the 5-sylow groups. Use that number to show that the subgroup must be normal

4. Aug 10, 2007

### djxl

Thanks for the quick help. I understand the solution now .

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