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what are the subgroups of $\mathbb{Z}_2 \times \mathbb{Z}_{12}$ of order $6$? I know that there are three such subgroups, and two subgroups are clear to me, namely the subgroup isomorphic to $\mathbb{Z}_6$ and the subgroup isomorphic to $\mathbb{Z}_2\times \mathbb{Z}_3$. But I can't see the other one. Please help!

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3 Answers 3

You can decompose $\mathbb{Z}_2\times \mathbb{Z}_{12}$ into $\mathbb{Z}_2\times \mathbb{Z}_4 \times \mathbb{Z}_3$. Clearly any group of order $6$ will contain $\mathbb{Z}_3$. Where else can you get your $2$ from?

Hint: I am guessing you've thought of getting the $2$ from $(1,0,0)$ and $(0,2,0)$. Why not both?

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Alternatively, any abelian group of order $6$ is cyclic, so you need to find all the elements of your group of order $6$ and then just realize that two of them generate the same subgroup if and only if they are the same or additive inverses.

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I think your "and then realize" needs more explanation if you want it not to be a rabbit from a hat. –  MJD Nov 3 '12 at 19:13
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The generators in an abelian group of order 6 come in pairs (other elements are identity or have order 2 or 3) - so the number of abelian subgroups of order 6 is half the number of elements of order 6. –  Mark Bennet Nov 3 '12 at 21:05
    
@MJD The cyclic group of order 6 has only two generators, and they are additive inverses. I was trying not to spell things out, however. –  Thomas Andrews Nov 4 '12 at 11:59
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Not sure what you mean by "isomorphic to $\mathbb Z_6$". All groups of order $6$ are isomorphic to $\mathbb Z_6$.

Here, the three subgroups of order $6$ are those generated by $(0,2)$, $(1,2)$, and $(1,4)$. Namely, $$ A_1=\{(0,0),(0,2),(0,4),(0,6),(0,8),(0,10)\}, $$ $$ A_2=\{(0,0),(1,2),(0,4),(1,6),(0,8),(1,10)\} $$ $$ A_3=\{(0,0),(1,4),(0,8),(1,0),(0,4),(1,8)\} $$

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