Group Algebras | Abstract Algebra

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A necessary condition for a group algebra to be reduced

Posted: November 25, 2022 in Group Algebras, Noncommutative Ring Theory Notes
Tags: dihedral group, group algebra, reduced ring

Throughout this post, $k$ is a field.

Let $G$ be a finite group. If $|G| \ge 2,$ then $k[G]$ is never a domain, a simple fact that we proved here (see Lemma 1). Now, let’s weaken the condition and ask whether or not $k[G]$ can have non-zero nilpotent elements. The answer this time is positive. For example, let $p$ be a prime number, $k$ a field of characteristic $p$ and $G$ a finite group such that $p \mid |G|.$ Let $g \in G$ be an element of order $p.$ Then $0 \ne 1-g \in k[G]$ is nilpotent because $(1-g)^p=1-g^p=0.$

A much more interesting question: for which fields $k$ and groups $G,$ does the group algebra $k[G]$ have non-zero nilpotent elements? That is not easy to answer in general but it is not hard to give a useful necessary condition for $k[G]$ to have no non-zero nilpotent elements, which is the subject of this post.

Recall that a ring is said to be reduced if it has no non-zero nilpotent element.

Lemma. Let $H$ be a finite subgroup of a group $G.$ If $k[G]$ is reduced, then $|H|$ is invertible in $k.$

Proof. That is obvious if the characteristic of $k$ is zero. So suppose that the characteristic of $k$ is $p > 0$ and assume, to the contrary, that $|H|$ is not invertible in $k,$ i.e. $|H|$ is zero in $k$ or, equivalently, $p \mid |H|.$ Then, since $p$ is a prime number, $H$ has an element $h$ of order $p.$ But then $(1-h)^p=1-h^p=0,$ contradicting our assumption that $k[G]$ is reduced. $\ \Box$

Proposition. Let $G$ be a group. If $k[G]$ is reduced, then every finite subgroup of $G$ is normal.

Proof. Let $H=\{h_1, \cdots , h_n\}$ be a finite subgroup of $G.$ Note that, by the Lemma, $n$ is invertible in $k$ and so

$\displaystyle x:=\frac{1}{n}\sum_{i=1}^nh_i \in k[G].$

Now,

$\displaystyle \begin{aligned}x^2=\frac{1}{n^2}\left(\sum_{i=1}^nh_i\right)^2=\frac{1}{n^2}\sum_{i=1}^nh_i\sum_{j=1}^nh_j=\frac{1}{n^2}\sum_{i=1}^n\sum_{j=1}^nh_ih_j=\frac{1}{n^2}\sum_{i=1}^n\sum_{j=1}^nh_j =\frac{1}{n^2}\sum_{i=1}^nnx=x.\end{aligned}$

So $x$ is an idempotent element of $k[G]$ and hence, since $k[G]$ is reduced, $x$ is central, by Remark 3 in this post. Thus $gxg^{-1}=x$ for all $g \in G,$ which gives

$\displaystyle \sum_{i=1}^ngh_ig^{-1}=\sum_{i=1}^nh_i.$

So for each $i,$ there exists a $j$ such that $gh_ig^{-1} =h_j \in H,$ proving that $H$ is normal. $\ \Box$

Example. Let $G:=D_6,$ the dihedral group of order $6.$ So $G$ is generated by two elements $a,b,$ where the orders of $a,b$ are two and three, respectively, and $ba=ab^{-1}.$ The subgroup of $G$ generated by $a$ is not normal in $G$ and hence, by the Proposition, $k[G]$ is not reduced. We can also prove that directly; just see, for example, that $(a+b-b^2-ab)^2=0$ and so $a+b-b^2-ab$ is a non-zero nilpotent element of $k[G].$

Exercise 1. Show that the converse of the above Proposition is not always true.

Exercise 2. Let $D_{2n}$ be the dihedral group of order $2n.$ Show that $k[D_{2n}], \ n \ge 3,$ always has a non-zero nilpotent element. What about $n=1,2$ ?

Locality of group algebras

Posted: August 14, 2022 in Group Algebras, Local & Semilocal Rings, Noncommutative Ring Theory Notes
Tags: augmentation ideal, group algebra, group of units, Jacobson radical, local rings, Maschke's theorem, p-group, ring of Laurent polynomials, semilocal rings

Note. The reference for the proof of the Theorem in this post is the first few pages of this paper. However, I have modified the proof given by the author of the paper quite a bit and also added all the details the reader needs to fully understand the proof.

Throughout this post, $k$ is a field, $\text{char}(k)$ is the characteristic of $k,$ and $G$ is a finite group. Also, as always, $J(R), U(R)$ denote, respectively, the Jacobson radical and the group of units of a ring $R.$ Finally, for a group algebra $R:=k[G],$ we denote by $\mathfrak{a}(R)$ the augmentation ideal of $R,$ i.e. $\mathfrak{a}(G)$ is the ideal of $R$ generated by the set $\{g-1: \ g \in G\}.$ See that $\mathfrak{a}(G)$ equals the $k$ -vector space spanned by $\{g-1: \ g \in G\}.$

In this post, we prove that the group algebra $k[G]$ is local if and only if $\text{char}(k)=p > 0$ and $G$ is a $p$ -group. This result is not very easy to prove; we need to do some preparation before getting into the proof.

Lemma 1. The group algebra $R:=k[G]$ is a domain if and only if $|G| =1.$

Proof. i) If $|G|=1,$ then $R \cong k$ and so $R$ is a domain. Now suppose that $|G| > 1$ and choose an element $1 \ne g \in G$ of order $d.$ Then $d > 1, \ g^d=1,$ and the elements $1, g, \cdots , g^{d-1}$ are pairwise distinct. Thus $1 - g \ne 0, \ 1+g + \cdots + g^{d-1} \ne 0$ but

$(1-g)(1+g+ \cdots + g^{d-1})=1-g^d=0,$

which proves that $R$ is not a domain. $\ \Box$

Lemma 2. For the group algebra $R:=k[G],$ the following statements are equivalent.

i) $R$ is local.

ii) $\mathfrak{a}(R) \subseteq J(R).$

iii) $\mathfrak{a}(R)$ is nil.

iv) $\mathfrak{a}(R)=J(R).$

Proof. i) $\implies$ ii). By Theorem 1 in this post, $R \setminus U(R)=J(R).$ Since $R \setminus U(R)$ contains every proper ideal of $R,$ we have $\mathfrak{a}(R) \subseteq J(R).$

ii) $\implies$ iii). Since $R$ is Artinian, $J(R)$ is nil, by the Example in this post, and so $\mathfrak{a}(R)$ is nil too. In fact $J(R)$ is nilpotent but we don’t need that here.

iii) $\implies$ iv). By the Remark in this post, $\mathfrak{a}(R) \subseteq J(R).$ By the Lemma in this post, $R/\mathfrak{a}(R) \cong k.$ Therefore $J(R)/\mathfrak{a}(R)$ is isomorphic to a proper ideal of the field $k$ hence $\mathfrak{a}(R)=J(R).$

iv) $\implies$ i). Again, by the Lemma in this post, $R/J(R) \cong k.$ Since $k$ is a field, hence a division ring, $R$ is local, by Theorem 1 in this post. $\ \Box$

Lemma 3. Let $G$ be a cyclic group of order $n.$ Then

i) $k[G] \cong k[x]/\langle x^n-1\rangle,$

ii) $k[G]$ is a (commutative) local ring if $\text{char}(k)=n.$

Proof. i) Let $g$ be a generator of $G.$ By definition, $\{1, g, \cdots , g^{n-1}\}$ is a $k$ -basis for $k[G].$ Define the map $f: k[x] \to k[G]$ by

$f(c_0+c_1x + \cdots c_mx^m)=c_0+c_1g+ \cdots + c_mg^m, \ c_i \in k.$

Clearly $f$ is a well-defined onto $k$ -algebra homomorphism. So we just need to show that $\ker f = \langle x^n-1\rangle.$ By definition, $f(x^n-1)=g^n-1=0$ and so $\langle x^n-1\rangle \subseteq \ker f.$ Conversely, let $p(x) \in \ker f.$ We have $p(x)=q(x)(x^n-1)+r(x)$ for some $q(x),r(x) \in k[x]$ with $r(x)=0$ or $\deg r(x) < n.$ If $r(x)=0,$ we are done. If $r(x)=\sum_{i=0}^m c_ix^i, \ c_i \in k, \ m < n,$ then

$0=f(p(x))=f(q(x))f(x^n-1)+f(r(x)) =f(r(x))=c_0+c_1g+ \cdots + c_mg^m$

giving $c_i=0$ for all $i.$ So again $r(x)=0.$

ii) Since $\text{char}(k)=n,$ we have in $k[x], \ x^n-1=(x-1)^n$ and so, by i), $k[G] \cong k[x]/\langle (x-1)^n \rangle,$ which is clearly a commutative local ring ring with the unique maximal ideal $(x-1)/\langle (x-1)^n\rangle. \ \Box$

Lemma 4. Let $Z$ be the center of $G$ and suppose that $N \subseteq Z$ is a subgroup of $G.$ If both $k[N]$ and $k[G/N]$ are local, then $k[G]$ is local too.

Proof. By the Corollary in this post, $k[G/N] \cong k[G]/I,$ where $I$ is the ideal of $k[G]$ generated by the set $\{g-1: \ g \in N\}.$ Note that since $N$ is central in $G,$ the ideal $I$ is central in $k[G].$ Since, by definition, $\mathfrak{a}(k[N])$ is the ideal of $k[N]$ generated by the set $\{g-1: \ g \in N\},$ we have $I=\mathfrak{a}(k[N])k[G].$ So, since $k[G/N]$ is local, $R:=k[G]/I$ is local too and hence $\mathfrak{a}(R)$ is nil, by Lemma 2. Now, let $x \in \mathfrak{a}(k[G]).$ Then $x+I \in \mathfrak{a}(R)$ and so $x^n \in I$ for some integer $n.$ Since $k[N]$ is local, $\mathfrak{a}(k[N])$ is nil, by Lemma 2, and since $k[N]$ is in the center of $k[G],$ the ideal $I=\mathfrak{a}(k[N])k[G]$ is also nil. Hence $x$ is nil, because $x^n \in I,$ and therefore $\mathfrak{a}(k[G])$ is nil, proving that $k[G]$ is local, by Lemma 2. $\ \Box$

Theorem. If $|G| > 1,$ then $R:=k[G]$ is local if and only if $\text{char}(k)=p > 0$ and $G$ is a $p$ -group.

Proof. Suppose first that $R$ is local. If $\text{char}(k)=0,$ then by Maschke‘s theorem $J(R)=(0)$ and so $R$ must be a division ring. But then $R$ would be a domain, contradicting Lemma 1. So $\text{char}(k)=p > 0.$ Now let $q$ be any prime divisor of $|G|$ and let $g \in G$ be an element of order $q.$ Since

$(1-g)(1+g+ \cdots + g^{q-1})=1-g^d=0,$

we have $1-g \notin U(R)$ and $1+g+ \cdots + g^{q-1} \notin U(R).$ Hence, since

$1-g^m=(1-g)(1+g+ \cdots + g^{m-1}),$

we have $1-g^m \notin U(R)$ for all positive integers $m.$ Since $R$ is local, $R \setminus U(R)$ is an ideal of $R$ and so

$\displaystyle q1_G=1+g+ \cdots +g^{q-1}+\sum_{m=1}^{q-1}(1-g^m) \notin U(R).$

But if $q \ne 0$ in $k,$ then $q1_G \in U(R)$ because $q^{-1}1_G$ would be the inverse of $q1_G.$ Therefore $q=0$ in $k,$ i.e. $p=\text{char}(k) \mid q,$ which gives $q=p.$ Thus $p$ is the unique prime divisor of $|G|$ and hence $G$ is a $p$ -group.
Conversely, suppose that $\text{char}(k)=p > 0$ and $G$ is a $p$ -group. We want to show that $k[G]$ is local. The proof is by induction over $|G|.$ If $|G|=p,$ then we are done by Lemma 3, ii). Now suppose that $|G| > p$ and let $Z$ be the center of $G.$ Since $G$ is a p-group, $|Z| \ge p.$ Let $N \subseteq Z$ be a subgroup of order $p.$ Clearly $|N| < |G|$ and $|G/N| < |G|.$ So, by our induction hypothesis, both $k[N]$ and $k[G/N]$ are local. Thus, by Lemma 4, $k[G]$ is local too. $\ \Box$

Exercise. Show that $k[\mathbb{Z}] \cong k[x,x^{-1}],$ the ring of Laurent polynomials. Conclude that $k[\mathbb{Z}]$ has infinitely many maximal ideals, and so it’s not even semilocal, never mind local.
Hint. To prove that $k[\mathbb{Z}]$ has infinitely many maximal ideals, note that $k[x,x^{-1}]$ is the localization of $k[x]$ by the set $\{1, x, x^2, \cdots \}.$

The augmentation ideal of a group ring

Posted: October 27, 2011 in Group Algebras, Noncommutative Ring Theory Notes
Tags: augmentation ideal, augmentation map, commutator subgroup, Hurewicz theorem

Let $R$ be a commutative ring with identity and let $G$ be a group. Recall that the group ring, also called group algebra, $R[G]$ is a free $R$ -module which is also a ring. As a free $R$ -module, $G$ is a basis for $R[G].$ So an element $x \in R[G]$ has the form $x=r_1g_1+ \cdots +r_ng_n,$ for some $g_i \in G, \ r_i \in R,$ and, since $G$ is a basis for the free $R$ -module $R[G],$ this representation of $x$ is unique if $g_i \ne g_j$ for all $i \ne j.$ Elements of $R$ are all central, i.e. they are in the center of $R[G],$ which is why we also call $R[G]$ a group algebra. In this post, we take a look at a very important ideal of $R[G]$ called the augmentation ideal.

Lemma. Let $R$ be a commutative ring and let $G$ be a group. We define the map $f : R[G] \longrightarrow R$ by

$f(\sum_{g \in G} r_g g)=\sum_{g \in G} r_g.$

Then $f$ is an onto ring homomorphism and hence $R[G]/\ker f \cong R.$ Also, $\{g - 1_G: \ g \in G, g \neq 1_G \}$ is a basis for the free $R$ -module $\ker f.$

Proof. Obviously $f$ is well-defined, additive and onto. Now, let $x = \sum_{g \in G} r_g g, \ y = \sum_{g \in G} s_g g.$ Then

$f(xy) = f(\sum_{g \in G} (\sum_{g_1g_2=g} r_{g_1}s_{g_2}) g)=\sum_{g \in G} \sum_{g_1g_2=g} r_{g_1}s_{g_2}=(\sum_{g \in G} r_g)(\sum_{g \in G} s_g)$

$=f(x)f(y).$

So $f$ is a ring homomorphism. Therefore $\ker f$ is an ideal of $R[G]$ and hence an $R$ -module. Now, $x = \sum_{g \in G} r_g g \in \ker f$ if and only if $\sum_{g \in G}r_ g = 0$ if and only if

$x = x - (\sum_{g \in G}r_g)1_G=\sum_{g \in G}r_g(g-1_G).$

Thus $\ker f,$ as an $R$ -module, is generated by the set $\{g - 1_G : \ g \in G \}.$ Then, obviously, the set $B=\{g-1_G: \ g \in G, g \neq 1_G\}$ still generates $\ker f.$ To show that $B$ is a basis for $\ker f,$ as an $R$ -module, we suppose that $\sum_{g \in B}r_g (g-1_G)=0.$ Then

$\sum_{g \in B}r_g g = (\sum_{g \in B}r_g)1_G.$

But $g \neq 1_G$ for all $g \in B,$ and so $r_g = 0$ for all $g \in B. \ \Box$

Definition. The ring homomorphism $f,$ as defined in the lemma, is called the augmentation map and $\ker f$ is called the augmentation ideal of $R[G].$

Hurewicz Theorem. Let $G$ be a group with the commutator subgroup $G'.$ Let $I$ be the augmentation ideal of $\mathbb{Z}[G]$ and consider $I$ as an additive group. Then $\displaystyle \frac{G}{G'} \cong \frac{I}{I^2}.$

Proof. Define the map $\displaystyle \varphi : G \longrightarrow \frac{I}{I^2}$ by $\varphi(g)=g-1_G + I^2.$ Clearly $\varphi$ is well-defined because $g - 1_G \in I$ for all $g \in G.$ Also, since $(g_1 - 1_G)(g_2-1_G) \in I^2,$ we have

$\varphi(g_1g_2)=g_1g_2-1_G + I^2 = g_1g_2 -1_G - (g_1-1_G)(g_2-1_G) + I^2=$

$g_1 - 1_G + g_2 - 1_G + I^2 = \varphi(g_1) + \varphi(g_2).$

Thus $\varphi$ is a group homomorphism. So $\varphi(g^{-1})=-\varphi(g)$ and therefore, since $I$ is an abelian group, we have $\varphi(g_1g_2g_1^{-1}g_2^{-1})=\varphi(g_1) + \varphi(g_1^{-1})+\varphi(g_2)+\varphi(g_2^{-1})=0.$ Thus $G' \subseteq \ker \varphi$ and hence we have a group homomorphism $\displaystyle \overline{\varphi}: \frac{G}{G'} \longrightarrow \frac{I}{I^2}$ defined by

$\overline{\varphi}(gG')=g - 1_G +I^2.$

Now, to show that $\overline{\varphi}$ is an isomorphism, we will find an inverse for it. Define the map $\psi : I \longrightarrow G/G'$ by $\psi(\sum_{g \in G} n_g(g-1_G))=(\prod_{g \in G} g^{n_g})G'.$ Note that since $g_1g_2G'=g_2g_1G'$ for all $g_1,g_2 \in G,$ the map $\psi$ is a well-defined group homomorphism. Now,

$\psi((g_1 - 1_G)(g_2-1_G))=\psi(g_1g_2 -1_G -(g_1-1_G)-(g_2-1_G))=g_1g_2g_1^{-1}g_2^{-1}G'$ $=G'.$

So $I^2 \subseteq \ker \psi$ because, by the lemma, the set $\{(g_1-1_G)(g_2-1_G): \ g_1,g_2 \in G\}$ generates the additive group $I^2.$ Hence the map $\displaystyle \overline{\psi} : \frac{I}{I^2} \longrightarrow \frac{G}{G'}$ defined by

$\overline{\psi}(\sum_{g \in G}n_g (g - 1_G)+ I^2) = (\prod_{g \in G} g^{n_g})G'$

is a well-defined group homomorphism. It is now easy to see that both $\overline{\varphi} \circ \overline{\psi}$ and $\overline{\psi} \circ \overline{\varphi}$ are identity maps. $\Box$

Center of group algebras

Posted: February 8, 2011 in Group Algebras, Noncommutative Ring Theory Notes
Tags: center of group algebras

Theorem. Let $k$ be a field and let $G$ be a finite group. Let $\mathcal{C}_1, \cdots , \mathcal{C}_m$ be the conjugacy classes of $G.$ Let

$\displaystyle c_i = \sum_{g \in \mathcal{C}_i} g \in k[G], \ \ \ \ \ 1 \leq i \leq m.$

Then the set $\{c_1, \cdots , c_m \}$ is a $k$ -basis for $Z(k[G]),$ the center of $k[G],$ and thus $\dim_k Z(k[G])$ is equal to the number of conjugacy classes of $G.$

Proof. $c_1, \cdots , c_m$ are linearly independent over $k$ because if $i \neq j,$ then $\mathcal{C}_i \cap \mathcal{C}_j = \emptyset.$ So we only need to show that $\text{span} \{c_1, \cdots , c_m \} = Z(k[G]).$ Let $x = \sum_{g \in G} x_g g \in k[G],$ where $x_g \in k$ for all $g \in G.$

Claim. $x \in Z(k[G])$ if and only if $x_g = x_{h^{-1}gh}$ for all $g,h \in G.$

Proof of the claim. Well, $x \in Z(k[G])$ if and only if $xh = hx$ for all $h \in G.$ Thus, $x \in Z(k[G])$ if and only if

$\displaystyle \sum_{g \in G} x_g g = \sum_{g \in G} x_g h g h^{-1}. \ \ \ \ \ \ \ \ (1)$

Clearly, fixing $h \in G,$ we have $\{hgh^{-1}: \ g \in G \} =G.$ So if we put $hgh^{-1}=u,$ then $g = h^{-1}uh$ and thus $(1)$ becomes

$\displaystyle \sum_{g \in G} x_g g = \sum_{g \in G} x_{h^{-1}gh} g. \ \ \ \ \ \ \ \ \ \ \ (2)$

The claim now follows from $(2).$

Now, if $1 \leq i \leq m,$ then, by definition, $g \in \mathcal{C}_i$ if and only if $h^{-1}gh \in \mathcal{C}_i$ for all $h \in G.$ So for every $g,h \in G,$ the coefficients of $g$ and $h^{-1}gh$ in $c_i$ are either both 1 or both 0. Thus, by the claim, $c_i \in Z(k[G])$ for all $i.$ So $\text{span} \{c_1, \cdots , c_m \} \subseteq Z(k[G]).$

Conversely, if $x = \sum_{g \in G} x_g g \in Z(k[G]),$ then, by the claim, $x_g = x_{g'}$ for all $g$ and $g'$ which are in the same conjugacy class. Thus if for every $i$ we choose a $g_i \in \mathcal{C}_i,$ then