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In this question I said that the automorphism group of $\mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/6\mathbb{Z}$ has 16 elements because

If $\varphi$ is one of this automorphism then $\varphi((1,0))=(1,0),(1,3),(3,3),(3,0)$ and $\varphi((0,1))=(0,1),(0,5),(2,1),(2,5)$.

They also said that $\mathrm{Aut}(\mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/6\mathbb{Z})\cong\mathbb{Z}/2\mathbb{Z}\times D_8$, and the proof seemed to work, but now I realized that $(0,1)$ can have 2 more images: $(2,2),(2,4)$ so it seems that the automorphisms are 24. What is the right answer?

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(2,2) = 2*(1,1) is a multiple of 2, but (0,1) is not, so though (2,2) has order 6, it is not a possible image of (0,1). This is basically happening since Z/4Z x Z/6Z = Z/12Z x Z/2Z, and you are taking the square of a generator of Z/12Z, which is not an independent generator. – Jack Schmidt Jan 30 '12 at 5:19
To add to Jack's comment, you want to map $(1, 0)$ and $(0, 1)$ such that the subgroups their images generate have the right size and intersect trivially. You found the elements of order $4$ in the group. It isn't too hard to see that the subgroups they generate will always intersect the group generated by $(2, 2)$ at the element $(2, 0)$. – Dylan Moreland Jan 30 '12 at 5:30
Derek Holt commented in your referred earlier question that automorphism group has order 16. If Derek writes it, I would believe it. – j.p. Jan 31 '12 at 10:19

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