[Math] Relationship between index $n$ subgroups and homomorphisms onto $\mathbb Z/ n \mathbb Z$

group-theory

I've been having a think about the relationship between index $n$ subgroups of a group $G$ and homomorphisms $G \to \mathbb Z / n \mathbb Z$.

Let's first have a look at the case $n = 2$.

i) Suppose we have a non-trivial homomorphism $\phi : G \to \mathbb Z / 2 \mathbb Z $, where $G$ is a group. Then $H = \mathrm{ker}\phi$ is a normal subgroup of $G$. Since $\phi$ is non-trivial, $G/H \cong \mathbb Z / 2 \mathbb Z$ and we have that $H$ is a normal subgroup of index $2$ in $G$. So every non-trivial homomorphism $\phi$ gives rise to an index $2$ subgroup $\mathrm{ker}\phi$.

ii) Suppose $H$ is a subgroup of index $2$ in $G$. Then $H$ is necessarily normal and not equal to $G$. It seems natural to define $\phi : G \to \mathbb Z / 2 \mathbb Z$ by $\phi(H) = 0$, $\phi(G \backslash H) = 1$. This is necessarily a homomorphism, since the product of two elements in $G \backslash H$ must be in $H$ (using normality of $H$). So every index $2$ subgroup gives rise to a non-trivial homomorphism, and the required correspondence is established.

Let's see what happens when we try to generalise this method for higher $n$.

If $n = 3$, any non-trivial homomorphism $\phi : G \to \mathbb Z / 3 \mathbb Z$ must be surjective, and so the argument used in i) still works; every non-trivial homomorphism $\phi$ gives rise to an index $3$ subgroup $\mathrm{ker}\phi$.

I then come across two problems:

  1. I can't seem to generalise i) for $n$ higher than $3$, since I can't guarantee surjectivity. EDIT: I can guarantee surjectivity if $n$ is prime, though (since non-triviality implies the existence of a $g \in G$ with $\phi(g) = a \neq 0$, and $n$ prime implies the existence of a multiplicative inverse of $a \ (\mathrm{mod } \ n )$).

  2. I can't even generalise my argument in ii) to the case when $n = 3 $, since normality was key here.

I suppose what I'd like to know is:

A) Is there a nice generalisation of the $n=2$ result? Does it use the same method and, if so, why haven't I managed to get it to work?

B) Is there some other approach to this problem / explanation of why no such approach exists? Perhaps it might be helpful to decompose $\mathbb Z / n \mathbb Z$ according to the prime factorisation of $n$?

C) Is there anything else relevant to the problem that I haven't thought about? I can see complications may arise in the case where $G$ is not finite.

Thanks.

Best Answer

The general statement of what you're heading toward is...

$G$ has a normal subgroup of prime index $p$ if and only if there exists a non-zero homomorphism from $G$ to $\mathbb{Z}_p$.

Proof: Suppose $H$ is a normal subgroup of index $p$. Then $G/H$ is a subgroup of order $p$. Thus it's isomorphic to $\mathbb{Z}_p$ since $p$ is prime: call this isomorphism $\varphi$. Let $\pi:G \to G/H$ be the projection homomorphism $\pi(g)=gH$. Then $\varphi \circ \pi$ is a non-zero homomorphism from $G$ to $\mathbb{Z}_p$.

Now suppose there is a non-zero homomorphism. Since any non-zero element of $\mathbb{Z}_p$ generates the group, this homomorphism is onto and so by the isomorphsm theorem $G/\mathrm{Ker}$ is isomorphic to $\mathbb{Z}_p$ which implies that the kernel (a normal subgroup) has index $p$.

Now if there is a surjective homomorphism onto $\mathbb{Z}_n$ then one has a normal subgroup of index $n$ (by the isomorphism theorem). However, if one has a normal subgroup of index $n$ (not nec. prime), then there is a surjective homomorphism onto a group of order $n$, but in general this group doesn't have to be cyclic (or even abelian).

Now if you want to probe subgroups of index $n$ which are not necessarily normal, your line of inquiry probably won't pan out -- because you're essentially looking at kernels (which are always normal). To look at subgroups (not nec. normal), group actions are the way to go.

We say $G$ acts on a set $S$ if there is a map $\cdot:G \times S \to S$ denoted $(g,s) \mapsto g \cdot s$ such that $(gh)\cdot s = g \cdot (h \cdot s)$ for all $g,h \in G$ and $s \in S$ and also $e \cdot s= s$ for all $s \in S$ if $e$ is the identity of $G$.

Let $s \in S$. Then $G \cdot s = \{ g \cdot x \;|\; g \in G\}$ is the orbit of $s$ and $G_s = \{ g \in G \;|\; g \cdot s = s \}$ is the isotropy subgroup fixing $s$ (or stabilizer of $s$). It's not hard to show that isotropy subgroups are in fact subgroups and the orbits of elements partition the set being acted upon. It can be shown that $[G:G_s]=|G \cdot s|$ (sort of a generalization of Lagrange's theorem), so the size of the orbit of $s$ times the size of the stabilizers of $s$ equals the size of the group.

Now suppose $H$ is a subgroup of $G$ of index $n$, then $G$ acts transitively on the left cosets of $H$ in $G$ (which is a set with $n$ elements): $g \cdot xH = (gx)H$. [Transitive means that there is only 1 orbit.]

Conversely, if $G$ acts transitively on a set of size $n$ then any isotropy subgroup (stabilizer) is a subgroup of index $n$ since the index of an isotropy subgroup fixing $x$ is equal to the size of the orbit of $x$ which is everything (i.e. $n$ elements) if we assume the action is transitive.

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