# Surjective Homomorphism

• Dec 27th 2013, 04:08 AM
topsquark
Surjective Homomorphism
Quote:

Let $Z_{36} = $. For which $a \in \mathbb{Z}$ does the map $\psi _a : Z_{48} \to Z_{36} : \overline{1} \mapsto x^a$ extend to a well defined homomorphism? Can $\psi _a$ ever be a surjective homomorphism?
In case the symbols aren't clear, $Z_{36} = \mathbb{Z}/36 \mathbb{Z}$, ie. residue classes mod 36. This one is written multiplicatively while $Z_{48}$ is written additively.

I handled the a part of the question: all a such that (a, 36) = 1 will do the trick.

But isn't $\psi _a$ already a surjective homomorphism? Or am I supposed to show the modular nature of the problem creates the surjection?

Thanks!
-Dan
• Dec 27th 2013, 10:23 AM
johng
Re: Surjective Homomorphism
Dan,
I think the notation chosen just obscures the problem. The attachment changes notation, and I hope the results are now clear.

Attachment 29965
• Dec 27th 2013, 12:51 PM
topsquark
Re: Surjective Homomorphism
Quote:

Originally Posted by johng
Dan,
I think the notation chosen just obscures the problem. The attachment changes notation, and I hope the results are now clear.

Attachment 29965

Thank you. That did clear things up.

-Dan
• Dec 28th 2013, 05:50 AM
Deveno
Re: Surjective Homomorphism
One can ask more generally, if it is possible to have a surjective homomorphism from a group G of order 48, to a group H of order 36.

Suppose it were, then for the kernel of the homomorphism, K, we must have:

|H| = |G|/|K|, which implies that 48 is a multiple of 36.

This implies that for any surjective homomorphism G-->G', where |G| = 48, that |G'| must be a DIVISOR of 48.

Now, if G is CYCLIC, then any homomorphism from G to some other group G' has cyclic image, and if x is a generator for G, and our homomorphism is, say f, then f(x) is a generator for f(G).

In this case, G' is cyclic of order 36, so its ONLY subgroups (one of each) are subgroups of order dividing 36:

1,2,3,4,6,9,12,18 and 36.

Since any homomorphic image of G (which is cyclic of order 48) must have order dividing 48, possible orders for a homomorphic image of G are:

1,2,3,4,6,8,12,16,24, and 48.

The common integers on these two lists are:

1,2,3,4,6 and 12, which are precisely the divisors of gcd(36,48) = 12.

If we write G' = <u>, which subgroups explicitly have these orders?

The subgroups generated by u36/k, where k is one of the divisors of 12 listed above, namely:

<u36> = {e} --order 1
<u18> = {e,u18} --order 2
<u12> = {e,u12,u24} --order 3
<u9> = {e,u9,u18,u27} --order 4
<u6> = {e,u6,u12,u18,u24,u30} --order 6
<u3> = {e,u3,u6,u9,u12,u15,u18,u21,u24,u27,u30,u33} --order 12

Finally, as far as "groupness" is concerned, any generator of a cyclic group is as good as any other, so to get a COMPLETE list of the integers that will work (mod 36), we need to include any generator of the cyclic subgroups.

(note that if G' = <u> = <v> then the map u-->v induces an automorphism of G' (an isomorphism of G' with itself), composing this automorphism with a homomorphism f:G-->G', will of course yield a different homomorphism G-->G').

From the (image) subgroup of order 1, we of course get 36 = 0 (mod 36). This is the 0-homomorphism (trivial homomorphism), which always exists between ANY two groups.
From the image subgroup of order 2, we get just 18 (a cyclic group of order 2 has a unique generator, the element of order 2).
From the image subgroup of order 3, we get the integers 12 and 24 (both non-identity elements are generators).

Since the rest of the orders are non-prime, we are only going to get $\phi(k)$ generators, so 2 each from the groups of orders 4 and 6 (yielding the integers 9,27,6 and 30) and 4 from the image subgroup of order 12 (3,15,21, and 33).

So our "final list" of integers a that will work in your problem is:

{0,3,6,9,12,15,18,21,24,27,30,33} (mod 36), which means that (since 3 divides 36), any multiple of 3 will work.

There is nothing wrong with johng's argument, it is elegant and thoroughly convincing, however: delving "into the details" reveals an interesting fact:

$12 = \sum_{k|12} \phi(k)$ (12 = 1 + 1 + 2 + 2 + 2 + 4)

something which can be seen to be true for ANY number (not just 12), by considering that every element of a finite cyclic group has SOME order, and the sum of the number of all elements of any given order over all possible orders must add up to the total number of elements in the cyclic group.

It turns out that cyclic groups of prime-power order are the "basic building blocks" of ALL finite abelian groups, something that is deeply linked with the multiplicative properties of prime numbers in the integers. In other words we can uniquely "factor" abelian groups in much the same way as we can factor integers, up to a (more or less) trivial rearrangement of the "factors" (put another way: the unique factorization property of the integers is a consequence of the Jordan-Holder theorem for finite abelian groups). This deep connection actually makes group theory very useful in number theory, questions about numbers can often be re-phrased in terms of finite abelian groups (like: divisibility).

This is no accident: the internal structure of abelian groups has a lot to do with integers, which make their influence felt in many places you might not expect. This exercise you are doing right now has a lot more to do with 36 and 48 than it has to do with groups, per se. A key role is played by 12 = gcd(36,48), and it turns out the number we are looking for is 36/12 = 3, which gives (generates) all the other possible answers.