# Automorphism of the Dihedral Group D8

• Dec 8th 2011, 12:26 PM
Bernhard
Automorphism of the Dihedral Group D8
Dummit and Foote Section 4.4 Automorphisms, Exercise 3 reads as follows:

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Prove that under any automomorpism of http://latex.codecogs.com/png.latex?D_8, r has at most two possible images and s has at most 4 possible images. Deduce that |Aut(http://latex.codecogs.com/png.latex?D_8)| http://latex.codecogs.com/png.latex?\leq8

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In a post on Project Crazy Project nbloomf gives a proof:

First the following Lemma is stated and proved:

Lemma: Let http://s0.wp.com/latex.php?latex=%5C...&fg=000000&s=0 be a group homomorphism. Then http://s0.wp.com/latex.php?latex=%5C...&fg=000000&s=0, with equality if http://s0.wp.com/latex.php?latex=%5C...&fg=000000&s=0 is bijective

Then the proof for the D&F exercise follows:

Let http://s0.wp.com/latex.php?latex=%5C...&fg=000000&s=0 be an automorphism of http://s0.wp.com/latex.php?latex=D_8...&fg=000000&s=0. Automorphisms must preserve order, so that http://s0.wp.com/latex.php?latex=%5C...&fg=000000&s=0 has order 4 and http://s0.wp.com/latex.php?latex=%5C...&fg=000000&s=0 has order 2. There are 2 elements of order 4 in http://s0.wp.com/latex.php?latex=D_8...&fg=000000&s=0, so that we have at most 2 choices for http://s0.wp.com/latex.php?latex=%5C...&fg=000000&s=0. Similarly, there are 5 elements of order 2 in http://s0.wp.com/latex.php?latex=D_8...&fg=000000&s=0. However, one of these is in http://s0.wp.com/latex.php?latex=Z%2...&fg=000000&s=0. By the lemma, we have at most 4 choices for http://s0.wp.com/latex.php?latex=%5C...&fg=000000&s=0

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Similarly, there are 5 elements of order 2 in http://s0.wp.com/latex.php?latex=D_8...&fg=000000&s=0. However, one of these is in http://s0.wp.com/latex.php?latex=Z%2...&fg=000000&s=0. By the lemma, we have at most 4 choices for http://s0.wp.com/latex.php?latex=%5C...&fg=000000&s=0

Can anyone please give a much more explicit proof of this assertion - why does one of the elements being in http://s0.wp.com/latex.php?latex=Z%2...&fg=000000&s=0 mean that we have at most 4 choices for http://s0.wp.com/latex.php?latex=%5C...&fg=000000&s=0???

Appreciate some help.

Peter
• Dec 8th 2011, 01:54 PM
Deveno
Re: Automorphism of the Dihedral Group D8
because Z(D8) = {1,r^2}. so the center has to map to itself. an automorphism is bijective, meaning s can't map to r^2, because r^2 already is mapped to r^2. so we have exactly 4 choices:

s-->s
s-->rs
s-->r^2s
s-->r^3s

homomorphisms preserve commutativity, if ab = ba in G, then φ(a)φ(b) = φ(ab) = φ(ba) = φ(b)φ(a) in H. so φ(Z(G)) is always contained in Z(φ(G)).

if φ is an isomorphism, then φ(Z(G)) has to BE Z(φ(G)), if you have a bijective map f:A-->B then f(A) = B, right?
• Dec 8th 2011, 01:59 PM
Bernhard
Re: Automorphism of the Dihedral Group D8
Thanks Deveno

Still struggling a bit ... when you say "so the center has to map to itself" - why ... because of the Lemma?

Peter
• Dec 8th 2011, 02:23 PM
Bernhard
Re: Automorphism of the Dihedral Group D8
Concerning the centre mapping to itself, I really need someone to prove explictly that $\displaystyle \phi$($\displaystyle r^2$) = $\displaystyle r^2$

Can anyone help?

I am assuming that $\displaystyle \phi$(1)= 1 is true because $\displaystyle \phi$ is a hmomorphism

Peter
• Dec 8th 2011, 02:54 PM
Gusbob
Re: Automorphism of the Dihedral Group D8
Quote:

Originally Posted by Bernhard
Concerning the centre mapping to itself, I really need someone to prove explictly that $\displaystyle \phi$($\displaystyle r^2$) = $\displaystyle r^2$

Can anyone help?

I am assuming that $\displaystyle \phi$(1)= 1 is true because $\displaystyle \phi$ is a hmomorphism

Peter

Let $\displaystyle G$ be a group, $\displaystyle \phi$ an automorphism on $\displaystyle G$. Suppose $\displaystyle a\in Z(G)$ and $\displaystyle g\in G$. Then $\displaystyle \phi (ab) = \phi (ba)$. Since $\displaystyle \phi$ is an automorphism it follows that $\displaystyle \phi(a)\phi(g)=\phi(g)\phi(a)$. You can see that the image of an element of the center commutes with every element of $\displaystyle G$ (since $\displaystyle \phi(g) \in G)$.

You may ask why couldn't $\displaystyle \phi(g)$ be in the center instead. The reason is that if $\displaystyle g,g'$ are both elements of $\displaystyle G$ but are not in the center, you know for sure that $\displaystyle \phi (gg')\not = \phi(g'g)$, so $\displaystyle \phi(g)\phi(g')\not = \phi(g')\phi(g)$.
• Dec 8th 2011, 03:15 PM
Deveno
Re: Automorphism of the Dihedral Group D8
we know that φ is a homomorphism (every automorphism is).

this means, in particular, for ANY g in D8, φ(r^2)φ(g) = φ(r^2g) = φ(gr^2) = φ(g)φ(r^2).

so whatever φ(r^2) is, it commutes with every φ(g), for every g in D8. but φ is an automorphism which means it is ONTO.

so every element of D8 IS φ(g) for some g. so φ(r^2) is in the center of D8. well, we only have two choices:

φ(r^2) = r^2, or φ(r^2) = 1, because those are the only two elements the center has.

but φ(1) = 1, so 1 is already taken. so φ(r^2) has to be r^2.