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Given a finite group $G$, its commutator subgroup $H$ of $G$ and a cyclic normal subgroup $N$ of $G$, I'm trying to show that $hn = nh$ for all $n \in N$ and $h\in H$.

That basically means that every commutator fixes any $n$. My progress so far seems hardly any progress at all; if anything I feel like I'm making things messier. But I'm just going to write them anyway.

My first thought was that since $hnh^{-1} \in N$, $hnh^{-1} = n^r$ for some $r$ and we can show $r$ must be 1, but this went nowhere.

My second approach was to write, without lost of generality, $h=aba^{-1}b^{-1}$. Then the result becomes $a^{-1}b^{-1}nba = b^{-1}a^{-1}nab$ for any two elements $a$, $b$ of $G$. I feel like this should either be trivial somehow or additionally complicated.

Also, I'm having difficulty figuring out how the cyclic-normality of N comes into play here.

Any help would be greatly appreciated!

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Note: It is not true that $g n g^{-1} = n$. For example, take the Dihedral group. –  Calvin Lin Jan 6 '13 at 3:11
    
@CalvinLin, it is true if $\,g\in G'\,$, as given in the question. –  DonAntonio Jan 6 '13 at 3:13
    
Sorry, I meant $h$. Thanks for the note, I'm editing. –  Benjamin Lu Jan 6 '13 at 3:14
    
Note that not all elements of $H$ are of the form $aba^{-1}b^{-1}$. –  Alex Becker Jan 6 '13 at 3:16

2 Answers 2

up vote 4 down vote accepted

This is a nice exercise...Hints

1) $\,\operatorname{Aut}(N)\,$ is abelian

2) Every inner automorphism of $\,G\,$ is, when restricted to $\,N\,$ , is an element of $\,\operatorname{Aut}(N)\,$

3) $\,\forall x,y\in G\,\,,\,[x,y]^{-1}=[y,x]\,$

Try now to do something with this and, if after thinking it over for a while you're still stuck, write back below as a comment.

Added on request: As noted, $\,\operatorname{Aut}(N)\,$ is abelian and if $\,\phi_g\,$ denotes the inner automorphism determined by $\,g\,$ , then $\,\forall\,g\in G\,\,\,,\,\,\text{then}\;\; \left.\phi_g\right|_N\,\in\operatorname{Aut}(N)$ . We show now that any basic commutator $\,[x,y]\in H\,$ centralizes any element $\,n\in N\,$ :

$$[x,y]n[x,y]^{-1}=[x,y]n[y,x]=x^{-1}y^{-1}xyny^{-1}x^{-1}yx=\left(\phi_{x^{-1}}\phi_{y^{-1}}\phi_x\phi_y\right)(n)=$$

$$\stackrel{\text{Aut}(N)\,\,\text{is abelian!}}=\left(\phi_{x^{-1}}\phi_x\phi_{y^{-1}}\phi_y\right)(n)=Id_N\circ Id_N(n)=n$$

and since the above is true for any generator of $\,H=G'=[G,G]\,$ then it is true for the whole group.

Second solution: Perhaps easier: for any subgroup $\,K\leq G\,$ , the map $$f:N_G(K)\to\operatorname{Aut}(K)\,\,,\,\,f(k):=\phi_k=\,\text{conjugation by}\,\,k$$

is a group homomorphism (with $\,\phi_k(x):=kxk^{-1}\,$), whose kernel is precisely $\,C_G(K)\,$ , and from here

$$N_G(K)/C_G(K)\cong T\leq\operatorname{Aut}(K)$$

In our case, we have $\,N\triangleleft G\Longleftrightarrow N_G(N)=G\,$ , so that we get $\,G/C_G(N)\cong T\leq\operatorname{Aut}(N)\,$ .

But $\,\operatorname{Aut}(N)\,$ is abelian, so that

$$G/C_G(N)\,\,\,\text{is abelian}\,\,\Longleftrightarrow G'\leq C_G(N)\;\;\;\;\;\;\square$$

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Thanks for the hints. So far I imagine you are hinting at the fact that with restriction to $N$, the group of conjugation by $h$ is abelian? So if we switch the order of conjugations it wouldn't change the result? If it is so I still have a hard time seeing how that leads to the desired result. –  Benjamin Lu Jan 6 '13 at 21:39
    
I've added some stuff to my answer. Read it there, @BenjaminLu –  DonAntonio Jan 6 '13 at 21:51

Since $N$ is normal, "conjugation by $g$" gives a group homomorphism $\phi$ from $G$ to $Aut(N)$, the group of automorphisms of $N$. Since $N$ is cyclic, this group of automorphisms is abelian. But you know that the commutator subgroup is the "universal" subgroup that quotients $G$ into something abelian. That is, if there's any group homomorphism from $G$ to an abelian group $A$, it always factors through the group $G/H$. Now I aks ya, what's the kernel of $\phi$?

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could you explain what you mean by "factor through $G/H$"? –  Benjamin Lu Jan 6 '13 at 21:20
    
also, I'm guessing the kernel of $\phi$ is the centralizer of $N$? I apologize if I appear a tad slow, it has been a while since I did group theory and even then I wasn't very good to start with. –  Benjamin Lu Jan 6 '13 at 21:29
    
A map from $G$ to some other group $A$ "factors through $G/H$" if its kernel contains $H$. This implies that the map descends to a map $G/H \to A$. –  Ted Jan 6 '13 at 22:00
    
(1) Generally, a map $f: G \to A$ "factors through" another group $L$ if there exists a pair of maps $l: G \to L$ and $a: L \to A$ such that $f = a \circ l$. (2) Yep, the kernel of $\phi$ is the centralizer of $N$. This is because if the conjugation action by $g$ fixes every element of $N$, $g$ must commute with every element of $N$. –  user54535 Jan 7 '13 at 15:22

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