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How can I compute $\mathrm{Aut}(\mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/6\mathbb{Z})$?

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

So it seems to me that this groups contains 16 elements, am i right? And if I'm right there is a way to understand to which group of cardinality 16 this is isomorphic to?

EDIT: $\mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/6\mathbb{Z}\cong\mathbb{Z}/8\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}$. And $\mathbb{Z}/8\mathbb{Z}$ and $\mathbb{Z}/3\mathbb{Z}$ are characteristic (am I right?), so $\mathrm{Aut}(\mathbb{Z}/8\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z})\cong\mathrm{Aut}(\mathbb{Z}/8\mathbb{Z})\times\mathrm{Aut}(\mathbb{Z}/3\mathbb{Z})\cong\mathbb{Z}/8\mathbb{Z}^*\times\mathbb{Z}/3\mathbb{Z}^*$ and this group has cardinality 8, right? Where is my mistake?

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Maybe this post can be useful: math.stackexchange.com/questions/27200/number-of-automorphisms-of-a-direct-produ‌​ct-of-two-cyclic-p-groups –  WLOG Jan 27 '12 at 7:27
Note that if $m$ and $n$ aren't relatively prime, it's not too hard to show that $\mbox{Aut}(\mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z})$ is never abelian. That might help. –  Isaac Solomon Jan 27 '12 at 7:33
You are correct in sayiong that the full automorphism group consists of the 16 elements that you have listed. I am not sure how to help you understand the group. It is nonabelian of exponent 4, and SmallGroup(16,11) in the small groups database. –  Derek Holt Jan 27 '12 at 8:22
Using GAP and StructureDescription gives $Aut(C_4 \times C_6) \cong C_2 \times D_8$. –  Mikko Korhonen Jan 27 '12 at 8:54
$\mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/6\mathbb{Z}$ is not isomorphic to $\mathbb{Z}/8\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}$: the right-hand group has an element of order 8, the left-hand group does not. –  Greg Martin Jan 27 '12 at 9:03

1 Answer 1

up vote 5 down vote accepted

$G = Z_4\times Z_6 = Z_4\times (Z_2\times Z_3)$ where $Z_4\times Z_2$ and $Z_3$ are characteristic in $G$. It's not hard to show that the automorphism group of a group that is the direct product of two characteristic subgroups is the direct product of the automorphism groups of the two characteristic subgroups.

As $\mathop{Aut}(Z_3)$ is easily dealt with (it gives you the factor $Z_2$) you are left with determining $\mathop{Aut}(Z_4\times Z_2)$.

$Z_4\times Z_2$ has four elements of order 4 (there are further 3 elements of order 2 and the neutral element). An automorphism of $Z_4\times Z_2$ permutes the four elements of order 4, but maps the inverse of each element of order $4$ on the inverse of its image [This trivial remark explains why the automorphism group is not the full symmetric group $S_4$.]. Even if only one element of order 2 is the square of a(ll) element(s) of order 4, the image of the elements of order 2 are determined by the images of the elements of order 4.

I guess you can finish off now.

PS: If you like geometry, label the four corners of a square with the four elements of order 4 where the opposing corner is labeled with the inverse. [Caution: m.k.'s $D_8$ is the dihedral group with $8$ elements, mathematicians not doing finite group theory call this group $D_4$.]

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I didn't understand why an automorphism of $Z_4\times Z_2$ maps each element of order 4 on its inverse. And why the images of the elements of order 2 are determined by the images of the elements of order 4. Could you explain this, please? –  Alex M Jan 27 '12 at 23:37
@AlexM: Sorry for the statement about the inverses. I wrote something different from what I intended. I'll correct it now. –  ego Jan 30 '12 at 11:13
@AlexM: Your first question should be answered with my correction [you couldn't understand it because what I wrote was wrong]. –  ego Jan 30 '12 at 11:20
@AlexM: Your second question (in your comment) can be answered by the following stronger statement about $Z_4\times Z_2$: Multiplication with any fixed element of order $4$ gives a bijection between the elements of order $4$ and the elements of order dividing $2$. –  ego Jan 30 '12 at 11:22

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