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If $G$ is virtually abelian, is it true that any quotient of $G$ is? If it is true can you give me a proof?

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up vote 6 down vote accepted

A group $G$ is virtually $X$, where $X$ is a group-theoretic property, if and only if there exists a subgroup $H$ of $G$ such that $H$ has property $X$ and $[G:H]$ is finite.

Proposition. Suppose $X$ is a property that is inherited to quotients (that is, if $G$ has property $X$, and $N\triangleleft G$, then $G/N$ has property $X$). If $G$ is virtually $X$, then any quotient of $G$ is virtually $X$.

Proof. Suppose that $G$ is virtually $X$, and let $N\triangleleft G$ be a normal subgroup. Let $H$ be a witness to the fact that $G$ is virtually-$X$.

Then $HN/N$ is a subgroup of $G/N$; since $HN/N \cong H/H\cap N$, and $X$ is inherited to quotients, it follows that $HN/N$ has property $X$. Thus, it suffices to show that $HN/N$ has finite index in $G/N$.

I claim that $[G/N: HN/N] = [G:HN]$. Indeed, if $\{g_i\}$ is a complete set of coset representatives for $HN$ in $G$, then it is easy to verify that $\{g_iN\}$ is a complete set of coset representatives for $HN/N$ in $G/N$.

Thus, $[G/N:HN/N] = [G:HN] \leq [G:H]$ (since $H\subseteq HN$); thus, since $[G:H]\lt\infty$, we have that $HN/N$ is a finite index subgroup of $G/N$ with property $X$, so $G$ is virtually $X$. $\Box$

Corollary. If $G$ is virtually abelian, then any quotient of $G$ is virtually abelian.

Proof. Any quotient of an abelian group is an abelian group. $\Box$

Elsewhere we noted the dual question: if $X$ is a property that is inherited by subgroups, and $G$ is virtually $X$, then any subgroup of $G$ is also virtually $X$.

Recall that a class of groups $\mathcal{X}$ is said to be a pseudovariety if and only if it is closed under subgroups, homomorphic images, and finite direct products; it is a variety if it is also closed under arbitrary (unrestricted) direct products. For example, the class of all abelian groups, or the class of all nilpotent groups of class $2$, or the class of all groups of exponent $n\gt 1$, are examples of varieties; the classes of all finite groups, the class of all nilpotent groups, and the class of all solvable groups are important examples of pseudovarieties that are not varieties.

Proposition: If $\mathcal{X}$ is a pseudovariety of groups, then the class $\mathbf{V}(\mathcal{X})$ of all groups that are virtually $\mathcal{X}$ is a pseudovariety.

Proof. Above we proved that if $\mathcal{X}$ is closed under homomorphic images, then so is $\mathbf{V}(\mathcal{X})$, and referenced a proof for closure under subgroups. If $G_1,\ldots,G_n$ are in $\mathbf{V}(\mathcal{X})$, let $H_1,\ldots,H_n$ be subgroups such that $H_i\in\mathcal{X}$ and $[G_i:H_i]\lt\infty$ for each $i$. Then $H_1\times\cdots\times H_n$ is in $\mathcal{X}$ (since $\mathcal{X}$ is a pseudovariety), and $$[G_1\times\cdots\times G_n:H_1\times\cdots\times H_n] = [G_1:H_1]\cdots[G_n:H_n],$$ so $H_1\times\cdots\times H_n$ is of finite index in $G_1\times\cdots\times G_n$. Thus, $G_1\times\cdots\times G_n$ is in $\mathbf{V}(\mathcal{X})$. $\Box$

Corollary. If $\mathcal{X}$ is a variety, then $\mathbf{V}(\mathcal{X})$ is a pseudovariety.

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Related to this question. – Arturo Magidin Apr 6 '12 at 21:31

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