54
$\begingroup$

It is suggested as an exercise in Serge Lang's book "Algebra" to show that the commutator subgroup $G^c$ of a group $G$ is a normal subgroup.

I'd like to do that but I am afraid I need help,

I think the first thing I need to figure out is how a general element in the commutator subgroup looks like, so that I can check that the defining condition for normality is satisfied.

That is, supposing for a moment that a general element in $G^c$ is denoted by $g$, I need to show that $aga^{-1} \in G^c,$ for all $a \in G$.

But here I get stuck, first because I am unsure how to write a general element in $G^c$ - a simple product in $G^c$ is of the form $xyx^{-1}y^{-1}aba^{-1}b^{-1}$ where $a,b \in G$. I cannot see a way to simplify this - I am sure there is one, but somehow I am blind today.

The second thing then is, even if one tries out the conjugation of a simple element like $xyx^{-1}y^{-1}$ in $G^c$, again not simplification offers itself easily I think .. what am I missing ?

An alternative would be to find a homomorphism of $G$ whose kernel is precisely $G^c$ - here I tried to think of this as a map $G \times G \to G$ but whatever I cook up is not a homomorphism.

Thanks for your hints !!

$\endgroup$
  • 3
    $\begingroup$ You don't need to figure out what a general element in the commutator subgroup looks like. The commutator subgroup is generated by commutators. Show that the property of "being a commutator" is invariant under conjuation (in fact it is invariant under all automorphisms). $\endgroup$ – Qiaochu Yuan May 6 '12 at 18:52
  • 41
    $\begingroup$ A "direct" argument: If $c$ is in the commutator subgroup, and $g \in G$, then $gcg^{-1}c^{-1}$ is a commutator, hence, by closure, $(gcg^{-1}c^{-1})c = gcg^{-1}$ is in the commutator subgroup. I saw this somewhere in Mathematics Magazine many years ago. $\endgroup$ – Ted May 6 '12 at 18:57
  • 13
    $\begingroup$ Ted's argument is from L. Myers, "Normality of the Commutator Subgroup" Math. Mag. 68 (1995), p. 49. The same slick method shows any subgroup containing the commutator subgroup is a normal subgroup. $\endgroup$ – KCd May 6 '12 at 19:55
  • $\begingroup$ @harlekin Every normal subgroup of $G$ is the kernel of a homomorphism $f$ from $G$ to some other group $H$. What happens to the commutator subgroup under $f$ if you take $H$ an abelian group? $\endgroup$ – Nicky Hekster May 6 '12 at 21:17
80
$\begingroup$

Denote the commutator of $a$ and $b$ by $a^{-1}b^{-1}ab = [a,b]$.

If $u$ is an element from the commutator subgroup, then $g^{-1}ug = u(u^{-1}g^{-1}ug) = u[u, g]$ .


Another approach: the commutator subgroup is defined to be the subgroup generated by the commutators, so every element of the commutator subgroup is of the form $$[a_1, b_1][a_2,b_2]\ldots[a_n, b_n].$$ It is enough to show that $g^{-1}[a,b]g$ is always in the commutator subgroup, because then

$$g^{-1}[a_1, b_1][a_2,b_2]\ldots[a_n, b_n]g = (g^{-1}[a_1, b_1]g)(g^{-1}[a_2,b_2]g)(g^{-1}\ldots g)(g^{-1}[a_n, b_n]g)$$

is a product of elements from the commutator subgroup. When $\phi$ is any homomorphism, we have $\phi([a,b]) = [\phi(a), \phi(b)]$. Since for any $g \in G$ the map $\phi$ defined by $\phi(x) = g^{-1}xg$ is a homomorphism, the result follows.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

If $c \in G'$ and $g \in G$, then also $[g, c] = gc g^{-1} c^{-1} \in G'$. Because G' is closed under products, we would also have $(gc g^{-1} c^{-1})c \in G'$. But $(gc g^{-1} c^{-1})c = gcg^{-1} \in G'$, so by definition $G'$ is a normal subgroup of $G$.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

To prove that $G^{c}$ is a normal subgroup of $G$, it suffices to prove that if $(g,[c,d])\in{G}\times{G^{c}}$, then $g[c,d]g^{-1}\in{G^{c}}$. But, $$g[c,d]g^{-1} = [gcg^{-1}, gdg^{-1}].$$ Since $G^{c}$ is comprised by finite products of commutators in $G$, for any $x\in{G^{c}}$, it must be the case that $gxg^{-1}\in{G^{c}}$ for any $g\in{G}.$

| cite | improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.