Commutator group in the center of a group

In summary, if ##H\triangleleft G## such that ##H\cap [G,G] = \{\,e\,\}\,,## then ##H \subseteq Z(G)##.
  • #1
AllRelative
42
2

Homework Statement


[G,G] is the commutator group.
Let ##H\triangleleft G## such that ##H\cap [G,G]## = {e}. Show that ##H \subseteq Z(G)##.

Homework Equations

The Attempt at a Solution


In the previous problem I showed that ##G## is abelian iif ##[G,G] = {e}##. I also showed that ##[G,G]\triangleleft G##.

***
I am unsure what ##(H\cap [G,G])## represents.
Is ##(H\cap [G,G])## = ##[H,H]##,
or is it equal to {##[x,y] \in H | x,y \in G##}??
***
I began my solution like this:
##\forall [a,b] \in (H\cap [G,G])##, we have that ##a^{-1}b^{-1}ab = e##
##\Rightarrow ab = ba##

something missing

##\Rightarrow H\subseteq Z(G)##.
 
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  • #2
AllRelative said:
I began my solution like this:
##\forall [a,b] \in (H\cap [G,G])##, we have that ##a^{-1}b^{-1}ab = e \Rightarrow ab = ba##
That's the wrong direction. You started with an element which is the identity and want to show what? That ##e\in Z(G)##?

Skip the direction. What can you say about ##g^{-1}hgh\,##?
 
  • #3
You're right, I started the wrong way haha.

So ##g^{-1}hg \in G## because of the normality of ##H## in ##G##.

##g^{-1}h^{-1}gh## = ##g'h## with ##g'=g^{-1}h^{-1}g##.
 
  • #4
AllRelative said:
You're right, I started the wrong way haha.

So ##g^{-1}hg \in G## because of the normality of ##H## in ##G##.

##g^{-1}h^{-1}gh## = ##g'h## with ##g'=g^{-1}h^{-1}g##.
So ##g^{-1}hg \in H## because of the normality of ##H## in ##G##.

##g^{-1}h^{-1}gh## = ##h'h## with ##h'=g^{-1}h^{-1}g##.
 
  • #5
fresh_42 said:
So ##g^{-1}hg \in H## because of the normality of ##H## in ##G##.

##g^{-1}h^{-1}gh## = ##h'h## with ##h'=g^{-1}h^{-1}g##.
Sorry I still don't see it...

We have an ##h' = g^{-1}h^{-1}g##.
We can write ##(g^{-1}h^{-1}g)h = [g,h] = h'h## which belongs to ##H##.

I assume we have to show somehow that ##h'h = hh'## which would make ##H## abelian and therefore a subset of ##Z(G)##.

But ##h'h = hh'##
##(g^{-1}h^{-1}g)h = h(g^{-1}h^{-1}g)##

I don't think we can much with that last expression...
 
  • #6
AllRelative said:
Sorry I still don't see it...

We have an ##h' = g^{-1}h^{-1}g##.
We can write ##(g^{-1}h^{-1}g)h = [g,h] = h'h## which belongs to ##H##.

I assume we have to show somehow that ##h'h = hh'## which would make ##H## abelian and therefore a subset of ##Z(G)##.

But ##h'h = hh'##
##(g^{-1}h^{-1}g)h = h(g^{-1}h^{-1}g)##

I don't think we can much with that last expression...
We already have ##[g,h] \in [G,H] \subseteq [G,G]## and ##[g,h]=h'h \in H##. This is ##[g,h]\in H \cap [G,G]=\{\,e\,\}\,.##
So ##[g,h]=g^{-1}h^{-1}gh=e##.

Everything needed is already here.
 
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Related to Commutator group in the center of a group

What is the commutator group of a group?

The commutator group of a group is the subgroup generated by all the commutators (elements of the form aba-1b-1) of the group. It is denoted by [G,G] and is also known as the derived subgroup of the group.

What is the center of a group?

The center of a group is the set of elements that commute with every element of the group. In other words, it is the set of elements that remain unchanged when multiplied by any element of the group. It is denoted by Z(G).

How are the commutator group and the center of a group related?

The commutator group is a subgroup of the center of a group. This means that all the elements of the commutator group also belong to the center of the group. However, the center may also contain elements that are not in the commutator group.

What is the significance of the commutator group in the center of a group?

The commutator group in the center of a group is important because it helps us understand the structure of the group. It can provide information about the normal subgroups and quotient groups of the group, and it is also used in the study of simple groups.

How is the commutator group in the center of a group related to the abelianization of the group?

The abelianization of a group is the quotient group obtained by dividing the group by its commutator subgroup. The commutator group in the center of a group is isomorphic to the commutator subgroup, and therefore, it is also isomorphic to the abelianization of the group. This means that the commutator group in the center of a group can be used to study the abelian structure of the group.

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