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The group $\mathbb{Z}_4\oplus \mathbb{Z}_{12}/\langle(2,2)\rangle$ is isomorphic to one of $\mathbb{Z}_8, \mathbb{Z}_4\oplus \mathbb{Z}_2, \mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_2$.

I have already determine the group is not isomorphic to $\mathbb{Z}_8$. I am having trouble understanding why it is not isomorphic to one of the other two. I found an answer that explains why it is not $\mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_2$.

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I have not worked with a cyclic element such as $\langle(2,2)\rangle$. That could be what is causing my confusion. I know that $\langle2\rangle = \{\dots, -2, 0, 2, 4, \dots\}$.

Any help is appreciated. Thanks.

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  • $\begingroup$ i know the order of the factor group is $8$ $\endgroup$ – rover2 Dec 7 '17 at 14:16
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    $\begingroup$ In this case, $\langle (2,2)\rangle = \{ (2k \bmod 4, 2k \bmod {12}) : k \in \mathbb{Z}\} = \{ (2k \bmod 4, 2k \bmod {12}) : 0 \leqslant k < 6\}$. Generally, $\langle a\rangle = \{ ka : k \in \mathbb{Z}\}$, and if $a$ has order $m$, that is equal to $\{ ka : 0 \leqslant k < m\}$. $\endgroup$ – Daniel Fischer Dec 7 '17 at 14:19
  • $\begingroup$ yes, i know the order of $\langle(2,2)\rangle = 6$, i understand why the elements of $\langle(2,2)\rangle$ are what they are now. so i know that the order of the factor group is $\displaystyle\frac{4\cdot12}{6}=8$. $\endgroup$ – rover2 Dec 7 '17 at 14:21
  • $\begingroup$ Okay. So what is the part you have trouble understanding? $\endgroup$ – Daniel Fischer Dec 7 '17 at 14:22
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    $\begingroup$ Isomorphisms preserve the order of elements. If $a$ has order $m$, and $\phi$ is an isomorphism, then $\phi(a)$ also has order $m$. So if two groups are isomorphic, for each $m$ both contain the same number of elements of order $m$. (This is a necessary condition, not sufficient, I think.) The class of $(0,1)$ has order $4$, but no element of $\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2$ has order $4$ (they all have order $1$ or $2$). $\endgroup$ – Daniel Fischer Dec 7 '17 at 14:42
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The order of $(2,2)$ in $\mathbb{Z}_4\oplus\mathbb{Z}_{12}$ is easily seen to be $6$, so the quotient group has $4\cdot12/6=8$ elements.

The order of $x=(0,1)+\langle(2,2)\rangle$ is indeed $4$, because $$ 2x=(0,2)+\langle(2,2)\rangle\ne0+\langle(2,2)\rangle $$ but $$ 4x=(0,4)+\langle(2,2)\rangle $$ and $(0,4)=4(2,2)$. This excludes $\mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_2$.

Now take $x=(a,b)$; then $$ 4x=(4a,4b)=(4a,4a)-(0,4(b-a))= 2a(2,2)+(b-a)(0,4)=2a(2,2)-4(b-a)(2,2) $$ Can you finish?

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Since you mention the word elimination, let us consider using the Smith normal form. You are taking the quotient of $\mathbb Z\oplus \mathbb Z$ by the subgroup generated by $a=(4,0)$, $b=(0,12)$ and $c=(2,2)$. Observe that these are not independent elements in $\mathbb Z\oplus \mathbb Z$, since we have $$3a+b-6c = 0$$

This allows us to discard $b$ from our generating set. Thus we want to understand the quotient by the subgroup generated by $c=(2,2)$ and $a=(4,0)$. This gives a matrix, which I will now put into a diagonal form by column and row operations:

$$\begin{pmatrix} 4 & 0 \\ 2 & 2\end{pmatrix}\to \begin{pmatrix} 0 & -4 \\ 2 & 2\end{pmatrix}\to \begin{pmatrix} 0 & -4 \\ 2 & 0\end{pmatrix}\to \begin{pmatrix} 2 & 0 \\ 0 & -4\end{pmatrix}\to \begin{pmatrix} 2 & 0 \\ 0 & 4\end{pmatrix}$$

This shows that the quotient is $\mathbb Z/2\oplus \mathbb Z/4$.

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