3

There is a question in my question paper (Q. No. 11 - Part B) which tells to verify whether

The automorphism group $\text{Aut} (\mathbb Z/2 \times \mathbb Z/2)$ is abelian.

I don't understand which group is meant here by $\mathbb Z/2 \times \mathbb Z/2?$ Is it $\mathbb Z_2\oplus\mathbb Z_2?$

user113578
  • 536
  • Yes, it is the same group. The $\times$ means that you take the product set with componentwise operations, and the product and direct sum of two abelian groups are the same. – Pablo Zadunaisky Dec 06 '13 at 14:34
  • @PabloZadunaisky: What about the symbol "/"? – user113578 Dec 06 '13 at 14:35
  • Some people use $\mathbb{Z}/2$ to be short for $\mathbb{Z}/2\mathbb{Z}$ which is isomorphic to $\mathbb{Z}_2$, as you suspected. – Dan Rust Dec 06 '13 at 14:36
  • The symbol $/2$ is just shorthand for $\mathbb{Z}/2\mathbb{Z}$. It's meant to denote that you're "modding out" by the even integers - i.e., setting every even integer to zero. –  Dec 06 '13 at 14:36
  • 2
    What do you mean the automorphism group is "abelian 12"? I ask because it is neither abelian nor of order $;12;$ ... – DonAntonio Dec 06 '13 at 14:46
  • 1
    Someone should post an answer to this question, as it is already answered by your comments. :-) – BIS HD Dec 06 '13 at 14:47
  • @DanielRust: I think this comes from the ring context where ${\bf Z}/2$ is shorthand for ${\bf Z}/(2)$, which is simply the quotient by the ideal generated by $2$. – tomasz Dec 06 '13 at 18:20
  • Hint: The group $\mathbb Z_2\oplus\mathbb Z_2$ has three non-zero elements, all of order two. And the sum of any two is the third. Therefore any permutation of those three elements is an automorphism. – Jyrki Lahtonen Dec 08 '13 at 10:43

0 Answers0