# Math Help - Prove <a> is subgroup of C(a)

1. ## Prove <a> is subgroup of C(a)

For any element a in G, a group, prove that <a> is a subgroup of C(a).

Proof:

Now $ =$ { $e, a, a^2, ... , a^{n-1}$} if $|a| = n.$ So e would be in <a> for the least possible n, implies that <a> is nonempty. If <a> = {e}, then <a> is already a subgroup of C(a), so I assume <a> do not equal to {e}.

Let $a^i$ and $a^j$ be in <a>, and consider $C(a) = \{g \in G : ga = ag\ \forall a\}$

Now $(a^i)(a^j) = a^{i+j}$ , which is in <a>, and $(a^i)^{-1} = a^{-i}$, which is in <a> as well.

Furthermore, $(a^i)(a) = a^{i+1} = a^{1+i} = (a)(a^{i})$, so $a^{i} \in C(a)$ , thus proves <a> is a subgroup of C(a).

Q.E.D.

Is that right?

For any element a in G, a group, prove that <a> is a subgroup of C(a).

Proof:

Now $ =$ { $e, a, a^2, ... , a^{n-1}$} if $|a| = n.$ So e would be in <a> for the least possible n, implies that <a> is nonempty. If <a> = {e}, then <a> is already a subgroup of C(a), so I assume <a> do not equal to {e}.

Let $a^i$ and $a^j$ be in <a>, and consider $C(a) = \{g \in G : ga = ag\ \forall a\}$

Now $(a^i)(a^j) = a^{i+j}$ , which is in <a>, and $(a^i)^{-1} = a^{-i}$, which is in <a> as well.

Furthermore, $(a^i)(a) = a^{i+1} = a^{1+i} = (a)(a^{i})$, so $a^{i} \in C(a)$ , thus proves <a> is a subgroup of C(a).

Q.E.D.

Is that right?
Where did you show that for any $i>0$ that there exits a $j \ge 0$ such that $a^{-i}=a^j$

RonL

3. The centralizer $C(a) = \{ g\in G | ga = ag \}$ contains all $\left< a \right>$. Because if $g \in \left< a \right>$ i.e. $g=a^k$ then $ga = a^k a = aa^k = ag$ implies $g=a^k \in C(a)$.

4. then would a not be an element of the center because a does not commute with everything? how would you show this?

5. Originally Posted by sfspitfire23
then would a not be an element of the center because a does not commute with everything? how would you show this?
You are talking to a vacum, all the participants in this thread have moved on.

Also quote the post you are refering to so we know exactly what you are refering to.

CB

6. Ah, sorry,

So, if <a>=C(a) of a non abelian group, G, then would the element a be in the center? that is, would $a\in Z(G)$. This is similar to a homework Q i have here at university in Canada

7. Originally Posted by sfspitfire23
Ah, sorry,

So, if <a>=C(a) of a non abelian group, G, then would the element a be in the center? that is, would $a\in Z(G)$. This is similar to a homework Q i have here at university in Canada
Not necessarily.

For instance, let's take an example for a symmetric group G=S_4.
(1,2,3,4) generates a subgroup <(1,2,3,4)> of S_4, which equals to $C_G((1,2,3,4))$. But the center of S_4 is {e}. Actually, a symmetric group S_n of n>=3 has a trivial center {e}.

8. would there be a non-symmetric group this could work for? A permutation?

9. ah, possibly if the group was non-abelian, the identity would be the only thing to commute with everything and thus an element a would not commute with everything and thus would not be in Z(G).

thoughts?