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Okay so I need to compute an automorphism on $\mathbb{Z}_2\times\mathbb{Z}_2$ using the fact that if $f\colon G\to H$ is an automorphism and $G=\langle K\rangle$, then $f$ is determined by where it takes members of $K$.

I think that $\mathbb{Z}_2\times\mathbb{Z}_2 = \langle(0,1),(1,0)\rangle$ under addition but I'm not sure how exactly to compute this automorphism. Is it as simple as constructing a Cayley table? Or do I just need to show how the kernels will compute all elements in $\mathbb{Z}_2\times\mathbb{Z}_2$? Or do I actually need to construct the function?

I'm not sure what the standard is here.

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@amWhy: It’s just part of the statement of the general fact that an automorphism $f:G\to H$ is determined by $f\upharpoonright K$ if $G=\langle K\rangle$. In this case $G=H=\Bbb Z_2^2$. –  Brian M. Scott Nov 7 '12 at 16:53
    
So you want an automorphism, $f$, $f: \mathbb{Z}_2\times\mathbb{Z}_2 \to \mathbb{Z}_2\times\mathbb{Z}_2$? Re: my question about H: got it, thanks for clarifying. –  amWhy Nov 7 '12 at 16:54
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3 Answers

up vote 1 down vote accepted

Yes, $\Bbb Z_2\times\Bbb Z_2$ is generated by $\langle 0,1\rangle$ and $\langle 1,0\rangle$; for convenience let me call these elements $a$ and $b$. The other non-identity element is $\langle 1,1\rangle$, which I’ll call $c$.

  • If $f:\Bbb Z_2\times\Bbb Z_2\to\Bbb Z_2\times\Bbb Z_2$ is an automorphism, it must take $a$ and $b$ to different non-identity elements of $\Bbb Z_2\times\Bbb Z_2$. (Why must they not be the identity? Why must they be distinct?)

Thus, to construct an automorphism $f$ of $\Bbb Z_2\times\Bbb Z_2$ you need only pick one of $a,b$, and $c$ to be $f(a)$ and another of them to be $f(b)$. For instance, you might decide to let $f(a)=b$ and $f(b)=c$. Now just use the properties of a homomorphism to compute $f$ in its entirety. For instance, $c=a+b$, so you must have $f(c)=f(a)+f(b)=b+c=a$, and of course $f(\langle 0,0\rangle=\langle 0,0\rangle$, which I’ll call $e$. You can display $f$ in tabular form:

$$\begin{array}{rcc} x:&e&a&b&c\\ f(x):&e&b&c&a \end{array}$$

(Of course you may prefer to use the ‘real’ identities of $e,a,b$, and $c$ instead of these short aliases.) Without too much work you can construct all of the automorphisms and write out their corresponding tables.

If you want to count the automorphisms without actually writing them all down, you’ll want not only the bullet point above, but also this one:

  • Every function from $\{a,b\}$ to $\Bbb Z_2\times\Bbb Z_2$ that sends $a$ and $b$ to distinct members of $\{a,b,c\}$ extends to a unique automorphism of $\Bbb Z_2\times\Bbb Z_2$. (Why?)

Thus, to count the automorphisms of $\Bbb Z_2\times\Bbb Z_2$ you need only count the injective functions from $\{a,b\}$ to $\{a,b,c\}$. How many ways are there to pick $f(a)$? Once you’ve done that, how many ways are there to pick $f(b)$? How many ways are there to pick both?

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In case you are interested in constructing or defining some $f$ to be an automorphism from $\mathbb{Z}_2\times\mathbb{Z}_2$, as opposed to constructing an automorphic group:

Using i. m. soloveichik's suggestion, you can construct an automorphism $f$ to act on the elements of $\mathbb{Z}_2\times\mathbb{Z}_2$ by multiplication with any one of the six invertible $2\times2$ matrices whose entries are all elements of $\mathbb{Z}_2$.

E.g., if you multiply $x \in \mathbb{Z}_2 \times \mathbb{Z}_2$ by $U$, where $U$ is the strictly upper triangular $2\times 2$ matrix whose non-zero values are all $1$, then $f(x) = Ux$ defines an automorphism. (In this example, the elements of $\mathbb{Z}_2 \times \mathbb{Z}_2$ would be represented in the form of two-by-one column vectors, for example, $(1, 0)^T$.

Suppose $$ U = \begin{bmatrix}% 1 & 1 \\ 0 & 1% \end{bmatrix}% \quad \text{and}\quad \mathbb{Z}_2\times \mathbb{Z}_2 = \langle(0,1),(1,0)\rangle = \left\langle \begin{bmatrix}% 0\\ 1 % \end{bmatrix} \begin{bmatrix}% 1\\ 0 % \end{bmatrix} \right\rangle = \left\{ \begin{bmatrix}% 0\\ 0 % \end{bmatrix} , \begin{bmatrix}% 0\\ 1 % \end{bmatrix} , \begin{bmatrix}% 1\\ 0 % \end{bmatrix} , \begin{bmatrix}% 1\\ 1 % \end{bmatrix}% \right\}.$$

Then $f: \mathbb{Z}_2\times \mathbb{Z}_2 \to \mathbb{Z}_2\times \mathbb{Z}_2\;,$ where $\;\;f(x) = Ux, \;\; x \in \mathbb{Z}_2\times \mathbb{Z}_2$

defines an automorphism on $\mathbb{Z}_2\times \mathbb{Z}_2$.

Since $\mathbb{Z}_2\times \mathbb{Z}_2$ is has only four elements, you can easily confirm that $f$ defines an automorphism on $G = \mathbb{Z}_2\times \mathbb{Z}_2$. E.g., $$U \begin{bmatrix}% 0\\ 1 % \end{bmatrix}% = \begin{bmatrix}% 1\\ 1 % \end{bmatrix}% \quad\quad U \begin{bmatrix}% 1\\ 0 % \end{bmatrix}% = \begin{bmatrix}% 1\\ 0 % \end{bmatrix}% \quad$$ etc.

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+1 How could I make this "<...>" bigger than the standard case? The same is for "{...}". –  B. S. Aug 10 '13 at 12:12
    
@Babak: Edited accordingly :) –  amWhy Aug 10 '13 at 13:16
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The automorphism group $G$ of $V=Z_2\times Z_2$ is the same as the group of invertible 2 by 2 matrices with entries in $Z_2$. $G$ is a group of order 6. You can view the action as multiplying a matrix by a column vector with entries in $Z_2$.

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