# Thread:

1. ## Is Centralizer subgroup?

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

2. 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?

3. 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.

4. 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.

5. 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.

6. 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.

7. 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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