Group Homomorphisms- kernel

• Oct 17th 2012, 10:58 AM
MAX09
Group Homomorphisms- kernel
I was working on a question on the no. of homomorphisms from S3 to Z/6Z.

My approach was as follows. if f is a homomorphism from G-> G', ker(f) = {g: f(g) = e'} should be a normal subgroup of G.

I recalled from a theorem that the only non-trivial normal subgroup of S3 is A3, the alternating group on 3 letters.
So I guessed that there could only be one homomorphism. I have two questions.

Can we count the homomorphism which has {e} has its kernel?
Would it not imply that it is also an isomorphism contradicting the fact that S3 is isomorphic to Z/6Z?
I doubt the contradiction because S3 cannot be isomorphic to Z/6Z because Z/6Z is in turn isomorphic to Z6, which is a cyclic group. But I guess S3 is not cyclic.
• Oct 17th 2012, 11:41 AM
johnsomeone
Re: Group Homomorphisms- kernel
Yes - your reasoning is correct as far as it goes.
Any homomorphism will produce a kernel, and that kernel will be a normal subgroup. The kernel can't be {e}, because otherwise, since those two groups have the same (finite) order, a kernel of {e} would correspond to a homomorphism that's actually an isomorphism - and because one group is cyclic and the other isn't, they can't be isomorphic. You're absolutely right. (Aside: Another way to quickly see that they're not isomorphic is to observe that one is abelian while the other isn't.)

You've looked at the case of the kernel being A3, and the case of the kernel being {e}, which you eliminated. There's another normal subgroup of S3 that you haven't considered: S3 itself. Which homomorphisms, if any, would have that as a kernel? (Hint: don't overthink this - it's easy.)

In trying to find the homomorphisms of S3 to Z6, you've broken it down according to:
1) Kernel = {e}: No homomorphisms.
2) Kernel = S3: ??? homomorphisms.
3) Kernel = A3: ??? homomorphisms.

That's a good start, but you'll still need to then examine which homomorphisms, if any, exist in those cases (well, only case 3 requires any work). Just because A3 is a normal subgroup of S3 does *not* guarantee that there's a homomorphism from S3 to Z6 having A3 as a kernel. Also, at the other extreme, there might be several different homomorphisms from S3 to Z6 having A3 as their kernel.

Ex: Is there a homomorphism from S3 to Z3 having kernel A3? No. If there were, then the image of that
homomorphism would be isomorphic to S3/A3, which is the group of order 2. But, by Lagrange's theorem, Z3 doesn't have a subgroup of order 2.

So your reasoning so far is excellent, but you've still got more work in front of you for the problem of finding the homomorphisms between those two groups.
• Oct 17th 2012, 02:56 PM
Deveno
Re: Group Homomorphisms- kernel
there a "flip side" to the first isomorphism theorem (which is what you are appealing to by considering ker(f)), that it just as important as ker(f), im(f).

that is, not only do we have to consider the normal subgroups N of G (to see what ker(f) might be), but we also have to see if G' contains a subgroup isomorphic to G/N.

that's another way to look at what quotient groups are: homomorphic images (f(G) < G' is what we get when we "factor ker(f) out").

since sgn:S3-->{-1,1} is an onto homomorphism, and {-1,1} is a cyclic group of order 2 (because all groups of order 2 are cyclic, since 2 is prime)

if we have a homomorphism f:S3-->Z6, the image must be a cyclic subgroup of Z6 of order 2.

now Z6 is cyclic, and cyclic groups are SPECIAL. in particular, cyclic groups have exactly one subgroup of any order dividing the order of the group.

so Z6 has just ONE subgroup of order 2, and being a subgroup of a cyclic group, it is also cyclic.

it is easy to see that this group is {0,3}. so if a homomorphism f:S3-->Z6 with ker(f) = A3 exists, it must be this one:

f(e) = 0
f((1 2)) = 3
f((2 3)) = 3
f((1 3)) = 3
f((1 2 3)) = 0
f((1 3 2)) = 0

now, one could verify for all 36 possible products a*b in S3 that f(a*b) = f(a) + f(b), or:

since we know there is an isomorphism h:S3/A3-->{1,-1}, we could prove that {1,-1} and {0,3} are isomorphic (let's call the isomorphism k),

in which case koh is an isomorphism S3/A3--->{0,3} (and we're done. why?).

*******

the point i'm trying to make here, is that verifying maps are homomorphisms "element-by-element" is inefficient. homomorphims have PROPERTIES, and we have nifty theorems about them we can use, which let us side-step any actual calculation (which is good, if you are lazy like me).

S3/{e} ---> H < Z6 leads to: H of order 6, H isomorphic to S3 and Z6, which can't be true since S3 and Z6 aren't isomorphic.
S3/A3 ---> H < Z6 leads to: H of order 2 (so H must be {0,3}), possible since 2 divides 6, and there is only one such H, so only one such homomorphism (which works because S3/A3 is cyclic, and any two cyclic groups of the same order are isomorphic).
S3/S3---> H < Z6 leads to: well, H must be a subgroup of Z6 of order 1...how many of these are there?

with that out of the way, we can simply count how many homomorphisms we found, and counting is easy.
• Oct 18th 2012, 10:13 AM
johnsomeone
Re: Group Homomorphisms- kernel
This approach to finding the homomorphisms between two groups depends on finding all the normal subgroups of one, and then, using whatever insights seem available, finding all the homomorphisms which have that normal subgroup as a kernel. This is rather hit or miss - it's not systematic at all. There is another approach, that, while also not completely systemaic, is still moreso than your approach.

Recall that in linear algebra, one of the most used, perhaps the most used, observations is that once you've defined a set map on just a basis of one vector space into another vector space, then by linear extension you've actually uniquely defined a linear map between those two vector spaces. One might say "If you've defined it on a basis, then you've defined it everywhere." Groups and their homomorphisms have a similar property, although it's more complicated.

This approach requires that you've seen "group presentations" - meaning groups defined in terms of generators and relations. If you're looking to find all the homomorphisms between G and H, and if G is given in terms of a presentation < P | R > ( = < generators | relations >), then a *consistent* mapping from just the set of generators P into H will define a unique homomorphism from G to H, in a way exactly analogous to the linear maps & vector spaces case of "if you've defined it on a basis, then you've defined it everywhere." The complication here is what "consistent" means. It means that if you have a set map from P to H, it can only be consistent with being a homomorphism between G and H if the relations R will hold as images in H. (I'll show a simple example at the bottom). But if you can find such a "consistent" set map from P to H, then you will have, by extension, defined a unique homomorphism from G to H. Proving that requires a bit of work (looking at free groups and such), but conceptually it should be clear, since what is a group homomorphism but an identification of the elements of G with some in H that "multiply like they do in G." Since the generators and relations in G tell you *everything* there is to know about "how things multiply in G", then once you've identified elements in H multiplying the same way, then you've found a "shadow" of G in H, i.e. a homomorphism. (FYI - my use of "consistent" in this paragraph isn't some technical definition, so far as I'm aware. It's just me trying to be descriptive.)

To use this approach, finding all group homorphisms from $\displaystyle S_3$ to $\displaystyle Z_6$ requires giving a group presentation $\displaystyle S_3$.

I'll give you that to get it started: $\displaystyle S_3 = <a, b \ | \ a^3 = 1, b^2 = 1, ab = ba^2 >$.

In terms of cycles & permutations, a = ( 1 2 3 ), b = ( 1 2 ).

$\displaystyle \text{Write } Z_6 = \{[0], [1], [2], [3], [4], [5] \}.$

$\displaystyle \text{So the "consistent" set maps } \phi \text{ from the generators }\{a, b \} \text{ of } S_3 \text{ to } Z_6 \text{ are those satisfying: }$

$\displaystyle \phi(a)^3 = \phi(1) = [0], \phi(b)^2 = \phi(1) = [0], \text{ and } \phi(a) \phi(b) = \phi(b) \phi(a)^2.$

$\displaystyle \text{Note that since } Z_6 \text{ is abelian, the last condition there implies } \phi(a) = [0].$

From there, can you enumerate all the "consistent" maps $\displaystyle \phi$?
You need to determine $\displaystyle \phi(a), \phi(b)$ in $\displaystyle Z_6,$ and have that they're consistent with the relations of $\displaystyle S_3.$
You already know that $\displaystyle \phi(a) = 0$, and there's a severe restriction on $\displaystyle \phi(b)$, namely, $\displaystyle \phi(b)^2 = [0].$
If you can do this, then you'll have produced the list of homomorphisms between those two groups.
Because Deveno has already completed this, you can check your discovered homorphisms with his results.

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Example:

(Before I begin, remember that presentations use non-abelian notation, and that can get confusing with these abelian groups. 1 means the identity, but so does [0]. I've used bracket notation for elements to tyt to alleviate this confusing situation. Also, see * at bottom.)

$\displaystyle \text{Let }Z_n = <a \ | \ a^n = 1> = \text{ the cyclic group of order } n. \text{ Find all homomorphisms from } Z_4 \text{ into } Z_{12}:$

$\displaystyle \text{(I'm going to write the elements of the *group* } (Z_{12}, +, 0) \text{ as } Z_{12} = \{[0], [1], [2], ... [10], [11] \}. \ )$

$\displaystyle \text{With } Z_4 = <a \ | \ a^4 = 1> \text{, if I define } \phi(a) = b \in Z_{12}, \text{ when will that be "consistent"?}$

$\displaystyle \text{Consistent means consistent with all the } Z_4 \text{ relations, so with } a^4 = [0].$

$\displaystyle \text{"} \phi \text{ is consistent with the relation } a^4 \text{" means } \phi(a)^4 = \phi(1) = [0] \in Z_{12}.$

$\displaystyle \text{(That uses that any potential group homomorphism must send the identity to the identity.)}$

$\displaystyle \text{It's hopefully easy to see that } \{ x \in Z_{12} \ | \ x^4 = [0] \} = \{ [0], [3], [6], [9] \}.$

$\displaystyle \text{(Also, see * at the bottom.)}$

$\displaystyle \text{Thus there are exactly 4 group homomorphisms, } \{ \phi_1, \phi_2, \phi_3, \phi_4 \}, \text{ from } Z_4 \text{ to } Z_{12},$

$\displaystyle \text{because there are exactly 4 "consistent" ways to define } \phi(a) \in Z_{12}, \text{ namely: }$

$\displaystyle \phi(a) \in A, \text{ so } \phi_1(a) = [0], \phi_2(a) = [3], \phi_3(a) = [6], \text{ and } \phi_4(a) = [9].$

$\displaystyle \text{If I now write } Z_4 \text{ as } Z_4 = \{[0], [1], [2], [3] \}, \text{ then } a = [1], \text{ and with this notation have:}$

$\displaystyle \phi_1([k]) = [0], \ \phi_2([k]) = [3k], \ \phi_3([k]) = [6k], \text{ and } \phi_4([k]) = [9k].$

$\displaystyle \text{Now } \phi_1 \text{ is obviously the trivial homomorphism. What are the others? I'll do } \phi_4 \text{ : }$

$\displaystyle \phi_4([0]) = [0], \phi_4([1]) = [9], \phi_4([2]) = [18] = [6], \phi_4([3]) = [27] = [3].$

$\displaystyle \text{To explicitly check, first note the consequences of }\phi_4([0]) = [0] :$

$\displaystyle \phi_4([0]) + \phi_4([x]) = \phi_4([0] + [x]) \ \forall \ [x] \in Z_4.$

$\displaystyle \text{Now check the rest, using abelian-ness to reduce the work: }$

$\displaystyle \phi_4([1]) + \phi_4([1]) = [9] + [9] = [6] = \phi_4([2]) = \phi_4([1] + [1]),$

$\displaystyle \phi_4([1]) + \phi_4([2]) = [9] + [6] = [3] = \phi_4([3]) = \phi_4([1] + [2]),$

$\displaystyle \phi_4([1]) + \phi_4([3]) = [9] + [3] = [0] = \phi_4([0]) = \phi_4([1] + [3]),$

$\displaystyle \phi_4([2]) + \phi_4([2]) = [6] + [6] = [0] = \phi_4([0]) = \phi_4([2] + [2]),$

$\displaystyle \phi_4([2]) + \phi_4([3]) = [6] + [3] = [9] = \phi_4([1]) = \phi_4([2] + [3]),$

$\displaystyle \phi_4([3]) + \phi_4([3]) = [3] + [3] = [6] = \phi_4([2]) = \phi_4([3] + [3])$

$\displaystyle \text{Thus } \phi_4 \text{ is a homomorphism.}$

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* Note that these groups are abelian, so "powers" are repeated additions: [7]^3 = [7] + [7] + [7] = [21] = [9] in $\displaystyle Z_{12}.$
So don't get that confused with [7]^3 = [7][7][7] = [-5][-5][-5] = -[125] = -[5] = [7] in the ring of residues mod 12.)
I'm is only considering the group. I'm not mentioning the ring of residues mod 12 at all.
• Oct 18th 2012, 02:18 PM
Deveno
Re: Group Homomorphisms- kernel

instead of using relations: equations of the form expression a in generators (x,y,z etc.) = expression b, it is often more convenient to use relators: elements defined to be the identity.

for example, in the presentation of S3 defined above, for the relation a3 = 1, we have the relator a3, and for the relation:

ab = ba2, we have the relator: aba-2b-1 (which is equal to: abab, or (ab)2).

so if we have a set of relators, f:G-->H is a homomorphism for G = <P|R> iff f(G) = <f(P)> and f(R) = {e}.

what we mean by G = <P|R> is that G is F(P)/<N(R)>, where F(P) is the free group generated by the set of generator P (the structure of this group only depends on the cardinality of P: in other words: "it doesn't matter what "letters" you use as symbols for the generators") and N(R) is the smallest normal subgroup of F(P) containing R.

one caveat: there may be several possible presentations for a given group G, and it is not always possible to even tell when two presentations give the same group. this is because free groups (on more than one generator, at least) have a very finely-patterned and rich internal structure: for example, the free group on two letters has a subgroup isomorphic to the free group on 3 letters!

nevertheless, this approach can save a great deal of calculation, especially for cyclic groups (which are all quotients of the free group on one generator, also known as "the integers" (well, ok, "isomorphic to the integers")). if G is cyclic, we just look for where the generator might be sent, and the order of the image of the generator must divide the order of the generator:

we cant send x in G = <x> and |x| = 12, to an element (and have a homomorphism) in another group G' of order 5,10 or 24, for example. the only possible values for |f(x)| for a homomorphism f are 1,2,3,4,6 and 12.

in the case of S3 to Z6, we have the two generators a and b, and |f(a)|, |f(b)| must divide 3 and 2, respectively. in Z6, we can't have:

|f(a)| = 3, |f(b)| = 2, or else 0 = f(e) = f((ab)2) = f(ab) + f(ab) = f(a) + f(a) + f(b) + f(b), and f(a) must be 2 or 4, and f(b) = 3 giving:

2 + 2 + 3 + 3 = 2 + 2 = 4 ≠ 0 or:

4 + 4 + 3 + 3 = 4 + 4 = 2 ≠ 0.

so one of |f(a)|,|f(b)| must be 1 (possibly both).

suppose f(b) = 0. then from f((ab)2) = 0, we get:

f(a) + f(a) = 0, which implies that |f(a)| = 1 or 2 (and in particular CANNOT be 3). since 2 does not divide 3, if f(b) = 0, f(a) = 0 as well.

otherwise, if |f(b)| = 2, we must have f(a) = 0 (since |f(a)| = 1), which works since:

f((ab)2) = f(a) + f(a) + f(b) + f(b) = 0 + 0 + f(b) + f(b) = f(b) + f(b) = 0 (because |f(b)| = 2).

so:

f(a) = 0, f(b) = 0 implies f(S3) = 0
f(a) = 0 f(b) of order 2 implies f(S3) = ???, ker(f) = ???
f(a) ≠ 0, f(b) ≠ 0, not possible.
• Oct 18th 2012, 06:04 PM
MAX09
Re: Group Homomorphisms- kernel
Deveno and johnsomeone.. i was working on your previous replies before i got another pair of replies.... still working.. just wanted to let you know that my being silent doesn't mean i'm not active... i'm working and will get back to you...

love ya both