Exercises #5 – Augmentation Ideal

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

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

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

[latex]
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
[/latex]

so [latex]a_e=-\sum_{g\neq e}a_g[/latex]. Hence,

[latex]
\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
[/latex]
[latex]\blacksquare[/latex]

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

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

[latex]
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
[/latex]

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

[latex]\blacksquare[/latex]

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

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

[latex]\blacksquare[/latex]

Corollary. [latex]G[/latex] is finitely generated if and only if [latex]\ker \varepsilon[/latex] is finitely generated.

Proof. Let [latex]S[/latex] be finite in the previous two exercises.

[latex]\blacksquare[/latex]

Let [latex]G[/latex] be a group with commutator subgroup [latex]G'[/latex]. Prove

[latex]\frac{G}{G’}\cong \frac{\ker\varepsilon}{(\ker\varepsilon)^2}[/latex]

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

[latex]\blacksquare[/latex]

Definition. The ideal [latex]\ker \varepsilon[/latex] is called the augmentation ideal of [latex]\mathbb{Z}G[/latex].

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