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How can we compute direct product of G with itself such that G=GL(2,2). We know that order of G is 6 and then the order of its direct product is 36. Since G is non-abelian, how can we describe the automorphism group of direct product of G?

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The automorphism group of $\operatorname{GL}(2,2) \times \operatorname{GL}(2,2)$ is called the wreath product $\operatorname{GL}(2,2) \wr S_2$ and has elements of the form $((g,h),\pi)$ where $(g,h) \in \operatorname{GL}(2,2) \times \operatorname{GL}(2,2)$ and $\pi \in S_2$. This is a group of order 72.

Multiplication is done in a natural way: $$((g,h),\pi) \cdot( (x,y),\sigma) = \begin{cases} ((gx,hy), \pi\sigma) & \text{if } \pi=() \\ ((gy,hx), \pi\sigma) & \text{if } \pi=(1,2) \end{cases}$$

In general, moving a permutation like $\pi$ past a tuple like $(x,y)$ uses $\pi$ to reorder the $n$-tuple, so $\pi=()$ leaves it alone and $\pi=(1,2)$ flips it to $(y,x)$.

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  • $\begingroup$ In the case of multiplying GL(2,2) n-times, then automorphism group is called the wreath product GL(2,2) wr S_n. right? $\endgroup$
    – Nil
    Feb 27, 2014 at 15:54
  • $\begingroup$ Yes, then the $n$-tuples really are $n$-tuples, and $\pi \in S_n$ moves them around in the natural way. $\endgroup$ Feb 27, 2014 at 15:56

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