Is Centralizer subgroup?

**tttcomrader** · September 18th 2007, 05:47 PM

Problem: Let G be a group, prove that the centralizers of G is a subgroup of G.

Proof: By definitions, the centralizers of G, $C(a) = \{ g \in G:ga=ag \} \forall a \in G$

Now, the identity element, e, has the property of $ea=ae$ , thus e is in C(a). So C(a) is not an empty set.

Assume C(a) contains more than just {e}, since if that is the case C(a) would be a subgroup.

Let x,y be in C(a), then: $xa=ax,ya=ay \forall a \in G$

Then $(xy)a=x(ya)=x(ay)=(xa)y=a(xy)$
Thus xy is in C(a).

Now, $(xa)^{-1}=(ax)^{-1}$
$x^{-1}a^{-1}=a^{-1}x^{-1}$

So x^{-1} is in C(a), thus C(a) is a subgroup of G.

Q.E.D.

Now, I am not sure if I have proven the last part correctly, that is, the inverse of x is in C(a), would anyone please have a look?

Oh, and the test is tomorrow morning, normally I would never ask for a free answer, as I would like to work it out myself if at all possible. But would anyone please give me the correct answer if I'm wrong as this problem might show up in the test? If you aren't comfortable with it, I fully understand, I really do appreciate the help I'm getting from here, thanks!

K

**ThePerfectHacker** · September 18th 2007, 06:02 PM

Originally Posted by tttcomrader

Problem: Let G be a group, prove that the centralizers of G is a subgroup of G.

Proof: By definitions, the centralizers of G, $C(a) = \{ g \in G:ga=ag \} \forall a \in G$

Now, the identity element, e, has the property of $ea=ae$ , thus e is in C(a). So C(a) is not an empty set.

Assume C(a) contains more than just {e}, since if that is the case C(a) would be a subgroup.

Let x,y be in C(a), then: $xa=ax,ya=ay \forall a \in G$

Then $(xy)a=x(ya)=x(ay)=(xa)y=a(xy)$
Thus xy is in C(a).

Now, $(xa)^{-1}=(ax)^{-1}$
$x^{-1}a^{-1}=a^{-1}x^{-1}$

So x^{-1} is in C(a), thus C(a) is a subgroup of G.

Q.E.D.

Everything else was perfect. The only thing you need to show is $x\in C(a)\implies x^{-1} \in C(a)$ .
This means,
$xa = ax$
Thus,
$a=x^{-1}ax$
Thus,
$ax^{-1}=x^{-1}a$ .
Q.E.D.

Which book you use?

**tttcomrader** · September 19th 2007, 12:55 PM

Oh, man, I didn't have a chance to look at your reply this morning.

But the test was easy, I think I miss one or two questions, should be an A.

Thanks.

Btw, we use "Contemporary Abstract Algebra" by Joseph A. Gallian.

**ThePerfectHacker** · September 19th 2007, 04:52 PM

Originally Posted by tttcomrader

Oh, man, I didn't have a chance to look at your reply this morning.

But the test was easy, I think I miss one or two questions, should be an A.

Thanks.

Btw, we use "Contemporary Abstract Algebra" by Joseph A. Gallian.

Remember you got an A all because of me.

**Blue Calx** · September 24th 2008, 03:15 PM

Originally Posted by tttcomrader

Then $(xy)a=x(ya)=x(ay)=(xa)y=a(xy)$
Thus xy is in C(a).

I was having a problem with this step. Is it ok to assume that the operation is associative? If so, then why? Proving closure is the only hang-up I was having on this problem.

**spoon737** · September 24th 2008, 03:48 PM

Yes. Since $a,x,y \in G$ , and all elements in $G$ are associative under it's operation, then any subset of $G$ automatically inherits the associativity property.

**Blue Calx** · September 24th 2008, 04:11 PM

Originally Posted by spoon737

Yes. Since $a,x,y \in G$ , and all elements in $G$ are associative under it's operation, then any subset of $G$ automatically inherits the associativity property.

Right, because associativity is a property of ALL groups in the first place. I forgot that detail!

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