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I know that the group $\mathbb{Z}_2\times\mathbb{Z}_4$ has:

1 element of order 1 (AKA the identity) 3 elements of order 2 4 elements of order 4

I'm considering the set of all automorphisms on this group, denoted $\operatorname{Aut}(\mathbb{Z}_2\times\mathbb{Z}_4)$.

I know that:

  • The automrphism needs to map the identity to the identity
  • The automorphism needs to preserve the orders of the elements in the group.

So by my watch, the 3 elements of order 2 are permutated, and the 4 elements of order 4 are permuted.

I just wish to count how many automorphisms there are. I am wuite confused. Is it $3\times4=12$. I have a feeling this is wrong.

Michael Hardy
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QuantumEyedea
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    You repeatedly used a forward slash where you needed a backslash. Also notice \operatorname{Aut}. ${}\qquad{}$ – Michael Hardy Apr 18 '15 at 04:33
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    This may help:http://math.stackexchange.com/questions/27200/number-of-automorphisms-of-a-direct-product-of-two-cyclic-p-groups – Arpit Kansal Apr 18 '15 at 05:47
  • Note that one element of order 2 is in both of the subgroups of order 4 and the other two are not. – Josh B. Apr 18 '15 at 22:18

1 Answers1

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Note that $G$ is generated by two elements of order $4$.

To count the possible automorphisms we can focus on the maps to these two generators. To the first generator we can map any of the $4$ elements of order $4$, giving $4$ possibilities.

To the second generator we don't want to map the same element of order $4$ as the first generator. And we don't want to map the other element of order $4$ that shares the same subgroup of order $4$. That is, to the other generator we need to map one of the elements of order $4$ in the other subgroup of order $4$. This gives $2$ possibilities for the second generator, for a grand total of $4\times2=8$ automorphisms.

Note: You can also do this by noting that $G$ is generated by an element of order $4$ and an element of order $2$, but you have to be careful when assigning the element of order $2$. As I mentioned in the comments one element of order $2$ is in both of the subgroups of order $4$ and the other two are not. So an automorphism can't take that one one element of order $2$ to one of the other two. In particular the subgroup generated by the first generator of order $4$ contains that one element of order $2$, so it can't be the second generator.

Josh B.
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