# Thread: Element of a Group Ring

1. ## Element of a Group Ring

Let $o(a)=n, \unlhd G$. Then, the question is does $\widehat{} = 1+a+...+a^{n-1} \in Z(RG)$ hold, for $R$ an arbitrary ring?

Clearly, it is enough to show that $\widehat{}g = g\widehat{} \forall g \in G$, that is, $g^{-1}\widehat{}g = \widehat{}$.

So, according to my notes, $g(1+a+...+a^{n-1})g^{-1} = 1+a^g+...+(a^{n-1})^g$, which I understand perfectly well. However, apparently this is just $\widehat{}$ because $ \unlhd G$. I don't get why that is true!

My thought is perhaps that the action of $g \in G$ on elements of a (finite) normal subgroup $N$ perhaps induces a unique element of $N$? That is to say, for $\{x_0, x_i,...,x_n\} = N \unlhd G, |N| \leq \infty, g \in G$, then ${x_i}^g \neq {x_j}^g$ for $i \neq j$?

Thanks in advance!

2. Originally Posted by Swlabr
Let $o(a)=n, \unlhd G$. Then, the question is does $\widehat{} = 1+a+...+a^{n-1} \in Z(RG)$ hold, for $R$ an arbitrary ring?

Clearly, it is enough to show that $\widehat{}g = g\widehat{} \forall g \in G$, that is, $g^{-1}\widehat{}g = \widehat{}$.

So, according to my notes, $g(1+a+...+a^{n-1})g^{-1} = 1+a^g+...+(a^{n-1})^g$, which I understand perfectly well. However, apparently this is just $\widehat{}$ because $ \unlhd G$. I don't get why that is true!

My thought is perhaps that the action of $g \in G$ on elements of a (finite) normal subgroup $N$ perhaps induces a unique element of $N$? That is to say, for $\{x_0, x_i,...,x_n\} = N \unlhd G, |N| \leq \infty, g \in G$, then ${x_i}^g \neq {x_j}^g$ for $i \neq j$?

Thanks in advance!
as you mentioned we have $1+a + \cdots + a^{n-1} \in Z(RG)$ if and only if $1+a+ \cdots + a^{n-1}=1+gag^{-1} + \cdots + ga^{n-1}g^{-1}, \ \ \forall g \in G. \ \ \ \ \ (1)$

now since $o(a)=n,$ we have $a^i \neq a^j,$ and hence $ga^ig^{-1} \neq ga^jg^{-1},$ for all $0 \leq i \neq j \leq n-1.$ thus, since $RG$ is a free ring, every $ga^ig^{-1}$

in the RHS of (1) must be equal to some $a^j$ in the LHS of (1). this means $$ has to be a normal subgroup of $G.$

conversely, if $$ is a normal subgroup of $G,$ then $\{ga^ig^{-1}: \ 0 \leq i \leq n-1 \}= gg^{-1}=$ and hence (1) holds.

3. Originally Posted by NonCommAlg
as you mentioned we have $1+a + \cdots + a^{n-1} \in Z(RG)$ if and only if $1+a+ \cdots + a^{n-1}=1+gag^{-1} + \cdots + ga^{n-1}g^{-1}, \ \ \forall g \in G. \ \ \ \ \ (1)$

now since $o(a)=n,$ we have $a^i \neq a^j,$ and hence $ga^ig^{-1} \neq ga^jg^{-1},$ for all $0 \leq i \neq j \leq n-1.$ thus, since $RG$ is a free ring, every $ga^ig^{-1}$

in the RHS of (1) must be equal to some $a^j$ in the LHS of (1). this means $$ has to be a normal subgroup of $G.$

conversely, if $$ is a normal subgroup of $G,$ then $\{ga^ig^{-1}: \ 0 \leq i \leq n-1 \}= gg^{-1}=$ and hence (1) holds.

That's much easier than I was anticipating. I do have one further question - why is RG a Free Ring? According to Wiki, Group Rings are Free Modules, things I know next to nothing about, but Free Rings are not mentioned...

4. Originally Posted by Swlabr

why is RG a Free Ring? According to Wiki, Group Rings are Free Modules, things I know next to nothing about, but Free Rings are not mentioned...
by "free ring" here i meant a free R-module which also has a structure of a ring. (it would be an algebra if R was commutative) so it's not anything official. i just made it up!