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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.

  • 0
    What is the statement of the "$n=2$ result" that you want to generalise?2011-12-28
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    The statement "$G$ has a [normal] subgroup of index $n$ if and only if $G$ has a nontrivial homomorphism onto $\mathbb{Z}_n$" holds for $n=2$, and more generally for $n=p$ a prime (in the normal case only), but is false in general. For example, $S_4$ has a normal subgroup of order $4$, hence of index $6$, but $S_4$ does not have $\mathbb{Z}_6$ as a homomorphic image. A possible generalization is to think of the cyclic group of order $2$ not as $\mathbb{Z}_2$, but as $S_2$, the permutation group. A normal subgroup of index $n$ gives a homomorphism into $S_n$.2011-12-28
  • 0
    That index 2 subgroups of a group $G$ correspond to non-trivial homomorphisms $G \to \mathbb Z / 2 \mathbb Z$, I suppose.2011-12-28
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    The statement “ *G* has a subgroup of index *n* if and only if *G* has a nontrivial homomorphism onto a cyclic group of order *n* ” holds only for *n* = 2. For instance, AGL(1,*p*) has a subgroup of index *p*, but no non-trivial homomorphism onto a cyclic group of order *p*.2011-12-28
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    In my opinion, the key feature you are missing is confusing cyclic groups with symmetric groups. A group has a proper subgroup of index at most *n* if and only if it has a non-trivial homomorphism into a symmetric group on at most *n* points.2011-12-28
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    @ArturoMagidin Thanks. How do we show the existence of a nontrivial homomorphism onto $\mathbb Z_p$ given a subgroup of index $p$? Why have you written "[normal]"?2011-12-28
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    The condition on normal subgroups of index $n$ and homomorphisms into cyclic group of order $n$ can be generalized to all cyclic numbers; see [here](http://math.stackexchange.com/questions/67407/group-of-order-15-is-abelian). But *only* to cyclic numbers. If $n$ is not a cyclic number, let $G$ be a noncyclic group of order $n$, and take $\{e\}$ as your subgroup of index $n$.2011-12-28
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    @jonathan: The first statement (with $n=2$) is true whether we include the word "normal" or not, because "subgroup of index $2$" implies normal. For index a prime $p\gt 2$ you need to include the word normal for the statement to hold. Simply note that $G/H$ is a group of order $p$, hence cyclic. The canonical projection gives the homomorphism.2011-12-28
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    I'm about to log off for several hours, if not the rest of the day. I'll come back and write an answer when I get a chance if this is still up in the air.2011-12-28
  • 0
    Great, thanks all.2011-12-28

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