# Exercises #5 – Augmentation Ideal

Definition. For any group $G$, the augmentation map of the group ring $\mathbb{Z} G$ is defined to be the canonical ring homomorphism defined by $\varepsilon(g)=1$.

#### Exercise 1.

Prove the elements $\{g-e: g\neq e\}$ form a $\mathbb{Z}$-basis for $\ker \varepsilon$.

#### Solution.

Let $I$ be the free $\mathbb{Z}$-module generated by $\{g-e: g\neq e\}$. Note $\varepsilon(g-e)=\varepsilon(g)-1=0$, so $I\subseteq \ker\varepsilon$. Conversely, if $\sum_{g\in G} a_g g\in \ker \varepsilon$, then

$0=\varepsilon(\textstyle \sum_{g\in G} a_g g)=\varepsilon(\sum_{g\neq e} a_g g+a_e e)=\sum_{g\neq e}a_g + a_e$

so $a_e=-\sum_{g\neq e}a_g$. Hence,

$\sum_{g\in G} a_g g = \sum_{g\neq e} a_g g + a_ee = \sum_{g\neq e} a_g g – \sum_{g\neq e}a_ge = \sum_{g\neq e} a_g (g – e)\in I$
$\blacksquare$

#### Exercise 2.

If $S$ is the set of generators for $G$, prove $\{s-e: s\in S\}$ generates $\ker \varepsilon$ as a left-ideal.

#### Solution.

Let $g-e\in \ker \varepsilon$ and let $I$ be the (left) ideal generated by $\{s-e:s\in S\}$. As $\{g-e: g\neq e\}$ forms a $\mathbb{Z}$-basis, it suffices to show $g-e\in I$. We proceed by induction on the length of $g$ in terms of $S$-products. The length 1 case is trivial, so suppose our claim is true for $i\geq 1$. For $g=s_1\cdots s_{i+1}$, we have

$g-e=s_1\cdots s_{i+1}-e=s_1\cdots s_{i+1}-s_1+s_1-e=s_1(s_2\cdots s_{i+1}-e)+s_1-e$

By the induction hypothesis, $s_2\cdots s_{i+1}-e\in I$, so $g-e\in I$.

$\blacksquare$

#### Exercise 3.

Suppose $\ker \varepsilon$ is generated (as a left-ideal) by a set $\{s-e: s\in S\}$ for some subset $S\subseteq G$. Prove that $S$ generate $G$.

#### Solution.

Let $H$ be the subgroup in $G$ generated by $S$. Let $f$ be the basis element in $\mathbb{Z}[G/H]$ corresponding to $H$. By definition, any $s\in S$ fixes $f$, so $(s-e)f=0$ implying $I$ annihilates $f$ (as every element of $I$ is a $\mathbb{Z}[G]$-linear sum of elements of the form $s-e, s\in I$). If $g\in G$, then $gf=(g-e+e)f=(g-e)f+f=f$, so $g\in H$, hence $H=G$.

$\blacksquare$

Corollary. $G$ is finitely generated if and only if $\ker \varepsilon$ is finitely generated.

Proof. Let $S$ be finite in the previous two exercises.

$\blacksquare$

#### Exercise 4.

Let $G$ be a group with commutator subgroup $G'$. Prove

$\frac{G}{G’}\cong \frac{\ker\varepsilon}{(\ker\varepsilon)^2}$

#### Solution.

Let $I=\ker\varepsilon$. Define $\Phi:G\to I/I^2$ by $g\mapsto g-e+I^2$ and $\Psi:I\to G/G'$ by $\sum a_g(g-e)\mapsto \prod g^{a_g}G'$. Using the identity $(g_1g_2-e)-(g_1-e)(g_2-e) = (g_1-e)+(g_2-e)$, $\Phi$ is a homomorphism. Moreover, using the first exercise, $I^2$ is generated by the set $\{(g_1-e)(g_2-e):g_1,g_2\in G\setminus \{e\}\}$, so using the same identity, $\Psi$ factors through $I^2$.
Note $\Phi(g^{-1})=-\Phi(g)$, so $\Phi$ factors through $G'$. It is easily verified that $\Phi\circ\Psi$ and $\Psi\circ\Phi$ are the identities.

$\blacksquare$

Definition. The ideal $\ker \varepsilon$ is called the augmentation ideal of $\mathbb{Z}G$.

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