5.3.7. We can show that Aut(Z2 x Z2) S3 in two ways. (a) One way is to show that

Question

5.3.7. We can show that Aut(Z2 x Z2) S3 in two ways. (a) One way is to show that any automorphism is determined by an invertible 2-by-2 matrix with entries in Z2, that there are six such matrices, and that they form a group isomorphic to S3. Work out the details of this approach. 5.4. GROUP ACTIONS AND GROUP STRUCTURE 255 Another way is to recall that Z2x Z2 can be described as a group with four elements e, a,b, c, with each nonidentity element of order 2 and the product of any two nonidentity elements equal to the third. Show that any permutation of {a, b, c) determines an automorphism and, conversely, any automorphism is given by a permutation of {a,b, c) (b)

Explanation / Answer

a)

Think of Z2 x Z2 as a 2-dimensional vector space over the (field) Z2. Then, Aut(Z2 x Z2) can be thought of as the linear transformations between Z2 x Z2 and itself whose determinants are nonzero. By thinking of these transformations via 2 by 2 matrices, there are only 16 to check.

Fortunately, only 6 of them have nonzero determinant. They are
[1 0] [1 0] [1 1] [1 1] [0 1] [0 1]
[0 1] [1 1] [0 1] [1 0] [1 0] [1 1].

Now, there are only two groups of order 6: Z_6 (abelian) or S_3 (non-abelian). Since it may be checked quickly that the matrix multiplication among these elements is NOT commutative, we have one conclusion: Aut(Z2 x Z2) is isomorphic to S3.

I hope that helps!

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