quaternions and Diheral group

**ux0** · October 26th 2009, 09:44 AM

Prove that the quaternions $\bold Q$ and the Dihedral group $D_8$ are non-isomorphic groups of order 8.

**clic-clac** · October 26th 2009, 10:50 AM

An isomorphism preserves the order of an element. Therefore if, for example, there are strictly more elements of order $2$ in $D_4$ than in $\bold Q,$ then theese groups are not isomorphic.

**ux0** · October 26th 2009, 06:25 PM

ok well i know that there is only one element in $\bold Q$ of order 2, which is $-I$ , and i know it is normal, how do I figure out how many subgroups of order 2 are in $D_8$ ?

**clic-clac** · October 27th 2009, 03:39 AM

The dihedral group of order 8 can be seen as the group of symmetries of the square. There are two diagonal symetries, which are involutions, therefore $D_8$ (with your notation) contains at least 2 elements of order 2; that is sufficient to prove what you want.

$D_8$ has exactly 5 elements of order 2 (two diagonal symmetries, 2 lateral symmetries and one central symmetry) and 2 elements of order 4 (one rotation of $\frac{\pi}{2}$ and its inverse).

**Swlabr** · October 27th 2009, 06:02 AM

Originally Posted by clic-clac

There are two diagonal symetries, which are idempotent...

Maybe I am missing something here, but the only idempotent in a group is the identity - an idempotent is an element $a$ such that $a^2=a$ . In a group we can cancel this to get that $a=1$ ...

**clic-clac** · October 27th 2009, 06:18 AM

You're not missing anything, I wrote idempotent thinking of involution... My bad

Thanks for pointing out the error.

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