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Problem

Find $\mathrm{Aut}(G)$, $\mathrm{Inn}(G)$ and $\mathrm{Aut}(G)/\mathrm{Inn}(G)$ for $G = D_4$

My Attempt

I let $D_4 = \{e, x, y, y^2, y^3, xy, xy^2, xy^3\}$

I found that $\mathrm{Inn}(G)$ consists of 4 bijective conjugation functions, namely $\{\phi_e, \phi_x, \phi_y, \phi_{xy}\}$. For $\mathrm{Aut}(G)$, I found that there are 12 automorphisms.

Here is the following link that relates to the problem I am doing.

Based on the solution:

Lemma: If $\alpha$ is an automorphism of group $G$ and $G$ has generators $x$ and $y$ with orders $n$ and $m$, respectively then $\alpha(x)$ and $\alpha(y)$ are also generators for $G$ with orders $n$ and $m$.

Proof: First, let us show that the orders agree. If $g \in G$ has order $n$, let $a = \alpha(g) \in G$ have order $m$. Then, $a^m = 1$ but by applying $\alpha^{-1} \in \mathrm{Aut}(G)$, we get $g^m = 1$. However, $g$ has order $n$, so $n$ must divide $m$. Similarly, $g^n = 1$ and applying $\alpha$, we note that $a^n = 1$, and so we conclude $n = m$.

Secondly, since any element of $G$ can be written as a product of $x$'s and $y$'s, and $\alpha$ is a surjective homomorphism, it follows that any element of $G$ can also be written as a product of $\alpha(x)$ and $\alpha(y)$, hence they generate $G$.

Using this Lemma (or a similar argument), we note that an automorphism of $D_4$ must send $y$ to $y, xy, x^2y$ or $x^3y$ and $x$ to $x$ or $x^3$. Any such pairing is possible, thus there are $2 \cdot 4 = 8$ such automorphisms...

(The notations that someone use are different from what I denote.)

The question I have is: Why are there 8 automorphisms? Shouldn't there be 12 automorphisms? Here is what I have:

$$e \mapsto e$$ $$x \mapsto \text{ either } \{x,y^2, xy^2\}$$ $$y \mapsto \text{ either } \{y, y^3, xy, xy^3\}$$

Then, there are $1 \cdot 3 \cdot 4 = 12$ automorphisms.

Any advices or comments?

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  • $\begingroup$ How about: you can't map $x$ to $y^2$? $\endgroup$ – user452 May 19 '13 at 1:26
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We know that $$D_4=\langle x,y|x^2=y^4=1, (yx)^2=1\rangle = \{e, x, y, y^2, y^3, yx, y^2x, y^3x\}$$

Since $x^2 = 1 = (yx)^2 = (yxy)x$, we see that $x = yxy = y^2xy^2 = y^3xy^3 = y^nxy^n$. From this we can see that $|x| = |yx| = |y^2x|=|y^3x|=2$. Indeed:

$$(y^nx)^2=(y^nxy^n)x=x^2=1\;.$$

In fact, $2$ is the smallest positive integer $m$ such that $(y^nx)^m = 1$. On the other hand $|y^3|=4$. So if any automorphism is defined then, according to the Lemma you cited, it must preserve orders; so our possibilities are as follows:

One of

$$\begin{eqnarray} x&\to&x\\ x&\to&yx\\ x&\to&y^2x\\ x&\to&y^3x \end{eqnarray} $$

and one of

$$\begin{eqnarray} y&\to&y\\ y&\to&y^3 \end{eqnarray} $$

(Of course, in all cases $1\to 1$.)

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  • $\begingroup$ Oh. I see. Another question: If I know those 8 bijective functions for Aut$(G)$ and 4 elements in Inn$(G)$, how do I find Aut$(G)$/Inn$(G)$? Is it by simply taking the cosets of the set? $\endgroup$ – NasuSama May 19 '13 at 14:53
  • $\begingroup$ Nice work here, Babak! $+\checkmark$ $\endgroup$ – amWhy May 20 '13 at 0:56
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To determine $\mathrm{Inn}(D_4)$,first observe that the complete list of inner automorphisms is $$\phi_{R_{0}},\phi_{R_{90}},\phi_{R_{180}},\phi_{R_{270}},\phi_{R_{H}},\phi_{R_{V}},\phi_{R_{D}},\phi_{R_{D'}}$$

Here $R_0,R_{90},R_{180},R_{270}$ are rotations of square. H and V denote the reflection of square with horizontal and vertical reflection and D and D' are denote diagonal reflection.Our job is to determine the no. of repetitions in this list.

Now u can check easily that $\phi_{R_{180}}=\phi_{R_{0}}$ also,$\phi_{R_{270}}=\phi_{R_{90}}$

similarly,$H=R_{180} V$ and $D'=R_{180} D$ so we have $\phi_{R_{H}}=\phi_{R_V}$ and $\phi_{R_D}=\phi_{R_D'}$

there are four inner automorphisms $$\mathrm{Inn}(D_4)=\{\phi_{R_0},\phi_{R_{90}},\phi_{R_H},\phi_{R_D}\}$$

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