1. ## conjugacy classes

Let $\displaystyle g_1, g_2, \text{...},g_r$ be the representatives of the conjugacy classes of the finite group G and assume that these elements pairwise commute. Prove that G is abelian.

2. Did you try anything or have any ideas yet? How far did you get?

3. Originally Posted by Tinyboss
Did you try anything or have any ideas yet? How far did you get?
i took arbitrary elements from the conjugacy classes represented by $\displaystyle g_i \text{ and } g_j$, viz, $\displaystyle r g_i r^{-1} \text{ and } s g_j s^{-1} \text{ where } r, s \in G$
these two should commute given that $\displaystyle g_i \text{ and } g_j$ commute.
when i work it out it comes down to saying that $\displaystyle r g_i r^{-1} \text{ and } s g_j s^{-1}$ commute iff $\displaystyle [(s^{-1} r) g_i (s^{-1} r})^{-1}]g_j = g_j[(s^{-1} r) g_i (s^{-1} r})^{-1}]$
Calling $\displaystyle s^{-1}r=h$ we have that $\displaystyle h g_i h^{-1}\text{ and } g_j$ should commute. here i am stuck.

4. Hmm.. well that's not the way to go. Just because representatives of classes commute, doesn't mean that all elements between the classes commute. For example, consider the classes $\displaystyle [(1\,2)],[(3\,4\,5)]\subseteq S_5$. There must be another way..

EDIT: Try this.

First, we need the fact that, for any finite group $\displaystyle G$ and subgroup $\displaystyle H$, $\displaystyle G\neq \bigcup_{a\in G}a^{-1}Ha$.

Now, we let $\displaystyle G$ act on itself by conjugation. Since it is assumed that the $\displaystyle g_i$ commute pairwise, we have $\displaystyle g_i^{-1}g_kg_i=g_k$ for all $\displaystyle 1\leq i,k \leq r$. That is, for each $\displaystyle k$, $\displaystyle g_1,\ldots ,g_r\in G_{g_k)$ (the stabilizer of $\displaystyle g_k$ under this group action).

Let $\displaystyle g\in G$ be arbitrary. Then there exist $\displaystyle i\in \{1,\ldots,r\},x\in G$ such that $\displaystyle g=x^{-1}g_ix$. Hence $\displaystyle g=x^{-1}g_ix\in x^{-1}G_{g_k}x$ for each $\displaystyle k$.

We have shown that, for any $\displaystyle g\in G$, we can find $\displaystyle x\in G$ so that $\displaystyle g\in x^{-1}G_{g_k}x$. It follows that $\displaystyle G=\bigcup_{x\in G}x^{-1}G_{g_k}x$ (for each $\displaystyle k$).

By the claim above, since $\displaystyle G$ is finite, this forces $\displaystyle G=G_{g_k}$ for each $\displaystyle k$. That is, the conjugacy class of $\displaystyle g_k$ is $\displaystyle \{g_k\}$. We know that this means $\displaystyle g_k\in Z(G)$ and, since these elements form a complete set of representatives for the conjugacy classes of $\displaystyle G$, $\displaystyle G=\{g_1,\ldots,g_r\}$, and the group is abelian.

5. Originally Posted by topspin1617
Hmm.. well that's not the way to go. Just because representatives of classes commute, doesn't mean that all elements between the classes commute. For example, consider the classes $\displaystyle [(1\,2)],[(3\,4\,5)]\subseteq S_5$. There must be another way..

EDIT: Try this.

First, we need the fact that, for any finite group $\displaystyle G$ and subgroup $\displaystyle H$, $\displaystyle G\neq \bigcup_{a\in G}a^{-1}Ha$.

Now, we let $\displaystyle G$ act on itself by conjugation. Since it is assumed that the $\displaystyle g_i$ commute pairwise, we have $\displaystyle g_i^{-1}g_kg_i=g_k$ for all $\displaystyle 1\leq i,k \leq r$. That is, for each $\displaystyle k$, $\displaystyle g_1,\ldots ,g_r\in G_{g_k)$ (the stabilizer of $\displaystyle g_k$ under this group action).

Let $\displaystyle g\in G$ be arbitrary. Then there exist $\displaystyle i\in \{1,\ldots,r\},x\in G$ such that $\displaystyle g=x^{-1}g_ix$. Hence $\displaystyle g=x^{-1}g_ix\in x^{-1}G_{g_k}x$ for each $\displaystyle k$.

We have shown that, for any $\displaystyle g\in G$, we can find $\displaystyle x\in G$ so that $\displaystyle g\in x^{-1}G_{g_k}x$. It follows that $\displaystyle G=\bigcup_{x\in G}x^{-1}G_{g_k}x$ (for each $\displaystyle k$).

By the claim above, since $\displaystyle G$ is finite, this forces $\displaystyle G=G_{g_k}$ for each $\displaystyle k$. That is, the conjugacy class of $\displaystyle g_k$ is $\displaystyle \{g_k\}$. We know that this means $\displaystyle g_k\in Z(G)$ and, since these elements form a complete set of representatives for the conjugacy classes of $\displaystyle G$, $\displaystyle G=\{g_1,\ldots,g_r\}$, and the group is abelian.
now that's quite something!
i agree, the proof is complete.
but i can understand the proof. there was no way that i would think that way.
how does all that come to ones mind?

EDIT: AND THIS TOO IS A SERIOUS QUESTION, AS SERIOUS AS THE THREAD ITSELF!!

6. Originally Posted by topspin1617
First, we need the fact that, for any finite group $\displaystyle G$ and subgroup $\displaystyle H$, $\displaystyle G\neq \bigcup_{a\in G}a^{-1}Ha$.
to prove this please check whether this is correct or not.
$\displaystyle g_1 H g_1^{-1} = g_2 H g_2^{-1}$

$\displaystyle \iff g_2^{-1} g_1 H (g_2^{-1} g_1)^{-1}=H$

$\displaystyle \iff g_2^{-1} g_1 \in N_G(H)$

$\displaystyle \iff g_1 N_G(H)=g_2 N_G(H)$

so maximum number of distinct non-identity elements which could be there in $\displaystyle \bigcup_{a\in G}a^{-1}Ha$ is $\displaystyle (|H|-1)|G:N_G(H)|$
now since $\displaystyle H \leq N_G(H)$ the proposition follows.