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According to this article,

A group homomorphism is a map $f:G \rightarrow H$ between two groups such that the operation is preserved: $f(g_{1}g_{2}) = f(g_{1})f(g_{2})$ for all $g_{1}, g_{2} \in G$, where the product on the left-hand side is in $G$ and on the right-hand side is $H$.

Question

Let G be a group and let $Aut(G)$ be the group of automorphisms of G.

(a) For any $g \in G$ define $\phi_{g} : G \rightarrow G$ by $\phi_{g}(x)=g^{-1}xg$. Check that $\phi_{g}(x)$ is an automorphism.

(b) Consider the map

$\Phi : G \rightarrow Aut(G)$

$g \mapsto \phi_{g}$

Check that $\Phi$ is a homomorphism.

My Attempt

I did the first part.

Now, I need to do the second part of this problem. Here is my attempt for the second problem.

Observe that if we evaluate $\Phi$ at $gh$ with arbitrary $x$, then:

$\Phi(gh) = \phi_{gh}(x) = (gh)^{-1}x(gh) = h^{-1}(g^{-1}xg)h = h^{-1}\phi_{g}(x)h = \phi_{h}(\phi_{g}(x))$.

The problem is that I'm stuck with this part. I think that the operation of that $\Phi$ would be the composition of two functions.

Another thought: It would not make sense to say that:

$\Phi(gh) = \Phi(g)\Phi(h)$

because,

$\Phi (gh) = h^{-1}g^{-1}xgh$

But

$\Phi(g) = g^{-1}xg$

$\Phi(h) = h^{-1}xg$

$\Phi(g)\Phi(h) = g^{-1}xgh^{-1}xh$

The problem is that G is not assumed to be abelian (Well, if it is, then the equality holds.). Then, we can't take the product of $\Phi$'s to match with $\Phi_{gh}$. If I'm right, the solution to the second part of this problem is unlike the one to the first part of this problem.

Any comments or suggestions?

NasuSama
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  • Please, make titles less subjective and more informative. – Pedro Apr 27 '13 at 19:43
  • You didn't happen to mean $\phi_g(x)=gxg^{-1}$ did you, conjugation by $g$? Then you would be able to show that $\Phi(g)=\phi_g$ is a homomorphism. As it stands, I don't think you can show what you want for a general group $G$. – Warren Moore Apr 27 '13 at 19:49
  • Actually, it's the function given based on the assignment I need to work on. Usually, the function is $\phi_{g}(x) = gxg^{-1}$, but my instructor gives us this $\phi_{g}(x) = g^{-1}xg$ which makes the problem seem more intriguing. – NasuSama Apr 27 '13 at 19:51
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    Intriguing perhaps, but sadly the problem is impossible as you've stated it. (Also, you need to add @NAME if you want people commenting here to be notified) – Warren Moore Apr 27 '13 at 19:59

4 Answers4

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What you have is an antihomomorphism, as you have shown. This is because you're using conjugation by $g^{-1}$.

You should actually be looking at the function $\Phi$ that maps an element $g\in G$ to a function, namely that who sends $x$ to $gxg^{-1}$. So we can write it as $g\mapsto \phi_g$ where $\phi_g(x)=gxg^{-1}$.

The question is whether this is an homomorphism. So take $h,g\in G$. Then $\Phi(hg)$ is a function that maps $x$ to $$(hg) x(hg)^{-1}=h(gxg^{-1})h^{-1}$$ But this can be seen to be the composition of the function $\phi_h$ that sends $x\mapsto hxh^{-1}$ with that which sends $x\mapsto gxg^{-1}$, because $$\phi_h\circ \phi_g(x)=h(g xg^{-1})h^{-1}$$ It follows that $$\Phi(hg)=\phi_{hg}=\phi_h\circ \phi_g=\Phi(h)\circ \Phi(g)$$ Remember that our group operation in $\rm Aut $ is composition of functions.

Pedro
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    The function defined by the OP is $\phi_g(x)=g^{-1}xg$, not $gxg^{-1}$. – Warren Moore Apr 27 '13 at 19:55
  • @WarrenMoore Yes, I added something. – Pedro Apr 27 '13 at 19:56
  • @PeterTamaroff Your response for the problem with the usual function is correct. However, for the given problem I have, when I checked with this Wikipedia article, it says that an antihomomorphism is a homomorphism. What do you think about this? Would you disagree with this? – NasuSama Apr 27 '13 at 20:05
  • @WarrenMoore Oh. I was confused between $Y$ and $Y^{op}$. I have a question: What is the difference between these notations? Is it that the operation is reverse for the group $Y$? – NasuSama Apr 27 '13 at 20:12
  • @NasuSama Yes, we define $a\times b=b\cdot a$ where $\cdot$ is the original product in $G$. – Pedro Apr 27 '13 at 20:13
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    @user50229 Where am I being rude? I don't think I have been rude, but I am sorry if that is what it felt like. I guess we can agree that any sane algebra instructor knows the difference between an homomoprhism and an antihomomorphism. – Pedro Apr 27 '13 at 20:20
  • @user50229 Where is he being rude from his response? I don't see this happening, including the one you thought is rude. Please don't get into too much "fight". – NasuSama Apr 27 '13 at 20:24
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Firstly I think you are confusing the function $\Phi$ with $\phi$. We have $\phi_g: x \mapsto g^{-1}xg$, but $\Phi: g \mapsto \phi_g$.

But actually using the function $\phi$ as defined, I think we get an antihomomorphism so that $\Phi(gh) = \Phi(h)\Phi(g)$. Using $\phi_g = gxg^{-1}$ would give the homomorphism.

user50229
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  • Haven't you just shown that $\Phi(gh)=\Phi(h)\Phi(g)$? Not that $\Phi(gh)=\Phi(g)\Phi(h)$? – Warren Moore Apr 27 '13 at 19:42
  • Really? Is that even true? Did I finish the work, or did I try that in other way around and not get the equality? – NasuSama Apr 27 '13 at 19:45
  • @Warren Yes you are right, I think the homomorphism is when $\phi_g = gxg^{-1}$. In this case I think $\Phi$ is an antihomomorphism. – user50229 Apr 27 '13 at 19:46
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    Yes I agree. The conjugation map is defined as $g\mapsto gxg^{-1}$, which isn't was is defined in the OP's question... – Warren Moore Apr 27 '13 at 19:48
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Put $\Phi:g\longmapsto \phi_{g^{-1}}$

Federica Maggioni
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I think that the main problem is the way the internal operation in $\operatorname{Aut}(G)$ is define.

Let $*:\operatorname{Aut}(G) \times \operatorname{Aut}(G) \to \operatorname{Aut}(G)$ be an internal operation in $\operatorname{Aut}(G)$, where $f*g:=g \circ f$. It's easy to show that $(\operatorname{Aut}(G), *)$ is a group.

Now, if we do the same reasoning, we get:

$$ \Phi(gh) = \phi_{gh}(x) = (gh)^{-1}x(gh) = h^{-1}(g^{-1}xg)h = h^{-1}\phi_{g}(x)h = \phi_{h}(\phi_{g}(x))=(\phi_{h}\circ\phi_{g})(x) $$

But then, we have that $(\phi_{h}\circ\phi_{g})(x)=\phi_{g}*\phi_{h}=\Phi(g)*\Phi(h)$. Which means that $\Phi$ is a group homomorphism.

Gerardo Vargas Flores
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