Originally Posted by

**jackie** Could someone help me on this problem?

I want to figure out which of the following groups is isomorphic to $\displaystyle (Z_4\times Z_{12}) / <(2,2)>$. The following groups are $\displaystyle Z_8, Z_2 \times Z_2 \times Z_2, Z_4 \times Z_2$

I know $\displaystyle <(2,2)>$ has order 6, so the given group has order 48/6=8. I tried to list the elements of this factor group which has order 8, but I was unable to do so since I don't know how to pick the representatives from $\displaystyle Z_4 \times Z_{12}$ so that they will give 5 distinct elements (I know $\displaystyle <(2,2)>$ is the identity in this factor group). I know $\displaystyle Z_8$ is cyclic, so if I can show that the factor group has no element of order 8, then I can eliminate $\displaystyle Z_8$. I also know every non-identity element of $\displaystyle Z_2 \times Z_2 \times Z_2$ has order 2, so if I can show that the factor group has an element of order 4, then I should be able to finish. Any help is appreciated.