Thread: Isomorphic Groups

Originally Posted by steph615

I need help determining in a, b, and c, which groups are isomorphic/not isomorphic to each other:

a) Z4 Z2 x Z2 P2 V
(V is the group of 4 complex numbers {i,-i,1,-1} with respect to multiplication)

b) S3 Z6 Z3 x Z2 Z7*

c) Z8 P3 Z2 x Z2 x Z2 D4

For (a), there are only two groups of order 4 (up to isomorphism): Z4 and Z2xZ2.
Note that Z4 is cyclic but Z2xZ2 is not.
V is cyclic (generated by i), so it must be isomorphic to Z4. I don't know what P2 is, I'm not sure what group you mean. But check if it's cyclic -- if it is, it's also isomorphic to Z4. If not, then it must be isomorphic to Z2xZ2.

For (b), there are again only two groups of order 6: Z6 and S3.
Note again that Z6 is cyclic, but S3 is not -- in fact, S3 isn't even abelian.
Z3xZ2 is abelian, so it cannot be S3. So it's isomorphic to Z6 (Note Znm is NOT always isomorphic to Zn x Zm, it only is when m and n are coprime. For instance, above, Z4 is not Z2 x Z2, but here Z6 is isomorphic to Z2xZ3)
And Z7* is cyclic (generated by primitive element 2), so also isomorphic to Z6.

For (c), there are five different groups to keep track of: 1 cyclic, 2 abelian noncyclic, and 2 nonabelian.
Z8 is the cyclic one, and none of the others are cyclic (barring P3 -- again I don't know what group the P's represent)
D4 is nonabelian (you may need to prove this).
Z2xZ2xZ2 is abelian and noncyclic. So far none of these are isomorphic to each other.
That leaves P3 which you'll have to do on your own, or give explanation about what that guy is.

Sorry, I should have explained what P was. P2 denotes the groups of subsets of a two –element set. So if D is a set, then the power set of D is the set PD of all the subsets of D.

Thanks for your help!