# almost complete need a few pointers to complete

Discussion in 'General Math' started by Abbey Vanderah, May 9, 2004.

1. ### Abbey VanderahGuest

I have this problem almost done but I'm not really sure how to finish
greatly be appreciated.

Question: Let G be a group. If |G|=30 and |Z(G)|=5, then determine the
isomorphsim class of G/Z(G).

Proof: Since G/Z(G) has order 30/5=6, there are two groups of order 6,
the cyclic group Z6, and the dihedral group S3. I think that the
isomorphism class of G/Z(G) is Z6 but I'm not really sure. Could
someone please explain to me how to get the isomorphism class.

Thanks for the help.

Abbey Vanderah, May 9, 2004

2. ### Brian VanPeltGuest

If G / Z(G) were isomorphic to Z6, then G / Z(G) would be cyclic
(since Z6 is cyclic).

There is a theorem that says if G / Z(G) is cyclic, then G is abelian.
This would imply that G = Z(G).

However, you have specified that Z(G) has order 5 and so Z(G) cannot
equal G. Hence, G is not abelian and G / Z(G) is not cyclic.

It looks to me like you are forced into G / Z(G) being S3.

Hope this helps,

Brian

Brian VanPelt, May 9, 2004