# Thread: Factor group

1. ## Factor group

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.

2. 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.

Writing $\displaystyle \mathbb{Z}_4=\{0,1,2,3\}\!\!\!\pmod 4\,,\,\,\mathbb{Z}_{12}=\{0,1,2,\ldots, 11\}\!\!\!\pmod{12}$ , we get $\displaystyle \mathbb{Z}_4\times \mathbb{Z}_{12}\geq <(2,2)>:=\{(2,2)\,,\,(0,4)\,,\,(2,6)\,,\,(0,8)\,,\ ,(2,10)\,,\,(0,0)\}$ .

Now note that $\displaystyle \forall\,b\in\mathbb{Z}_{12}\,,\,\,4b=0,4,8\!\!\!\ pmod{12}$ , so for any element $\displaystyle (a,b)$ in the direct product we get $\displaystyle 4(a,b)$ is one of $\displaystyle (0,0)\,,\,(0,4)\,,\,(0,8)$ ,and all these

elements belong to $\displaystyle <(2,2)>\Longrightarrow$ the maximal order of an element in the factor group is 4...does this now give you an idea what this factor group is?

Tonio