Heres the problem:

let

be a non-abelian group of order 8

1. prove there exists an element of

, call it

such that it's order is 4.

2. Let

be an element of

such that

is not formed by

. Show that

.

3. Show that if the order of

is 2 then

is isomorphic to

, and if the order of

is 4 then

is isomorphic to the quatrernion group.

My solution so far:

I was able to show 1 easily by using legrange's theorem and eliminating the possiblities of 8,1,2 being the only order of elements.

Since

is of order 4 it is obvious why

.

To show 2 I showed that the elements of

must be

. Then my idea is to show by elimination that

.

I was able to eliminate all options besides

and this is where I'm stuck.

contradiction.

Tonio
I'm not sure if this is the right approach at all or is there a much simpler and elegant way of showing this.

Thanks,

SK