Let G be a Klein Four Group. How many distinct homomorphisms $\displaystyle \phi : G \to G$ are there? How many of these are isomorphisms?

Stuck on how to start this. Any hints?

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- Feb 21st 2010, 10:50 AMnmatthies1Klein Four Group
Let G be a Klein Four Group. How many distinct homomorphisms $\displaystyle \phi : G \to G$ are there? How many of these are isomorphisms?

Stuck on how to start this. Any hints? - Feb 21st 2010, 11:18 AMtonio
- Feb 22nd 2010, 01:04 AMnmatthies1
I don't understand...What exactly do you mean by "is uniquely determined by its values on the generators http://www.mathhelpforum.com/math-he...09167892-1.gif, ..." ?

- Feb 22nd 2010, 01:57 AMHallsofIvy
The Klein four group has

**two**generators, a and b. Any homomophism must map generators to generators- it must map {a, b} into itself- or map one or more generators onto the identity (I added this last part- I**think**we need to include this). Any**isomorphism**must map the set of generators one-to-one and onto itself. - Feb 22nd 2010, 02:05 AMtonio

I think there is some confusion here: any HOMOMORPHISM from G to itself (i.e., an endomorphism of G) is uniquely determined by its action on any set of generators of G, but it doesn't HAVE TO map generators to generators. It can map a generator to an element whose order is a divisor of the generator's one, for example.

Now, an ISOMORPHISM from G to G (i.e., an automorphism of G) has to map generators to generators.

So, for example, the map $\displaystyle f(a)=b,\, f(b)=1,\,f(ab)=b,\,f(1)=1$ is an endomorphism of G but not an automorphism (why?), whereas the map $\displaystyle g(a)=ab,\,g(b)=a,\,g(ab)=b,\,g(1)=1$ is an automorphism of G(why?)

Tonio - Feb 22nd 2010, 03:34 AMTopologistDescription of the endomorphisms of the Klein group
Let us denote the Klein group by $\displaystyle G=\{e,a,b,c\}$, where

*e*is the identity of*G*and*a*,*b*and*c*are the three elements of order 2 in*G*. Intuitively, we cannot "distinguish" between*a*,*b*and*c*in a group-theoretic manner; notice that $\displaystyle a^2=b^2=c^2=e$, that $\displaystyle ab=ba=c$, that $\displaystyle ac=ca=b$, and that $\displaystyle bc=cb=a$. In some sense, therefore, there is a "symmetry" between the elements*a*,*b*and*c*.Mathematically, we say that there is an isomorphism of*G*that carries*a*to*b*, an isomorphism that carries*a*to*c*, an isomorphism that carries*b*to*c*; succintly, every permutation of*a*,*b*and*c*corresponds to an isomorphism of*G*. Therefore, there are precisely 6 isomorphisms from*G*to*G*, since there are precisely 6 permutations on three letters.

How do we determine the number of homomorphisms from*G*to*G*? Well, we can consider the kernel of each homomorphism in doing so (recall the first isomorphism theorem?). Since*G*is abelian, every subgroup of*G*is normal, and therefore a kernel for some homomorphism defined on*G*. However, if we deal exclusively with non-trivial homomorphisms that are*not*isomorphisms, the only possibilities for the kernel of the given homomorphism are $\displaystyle K_1=\{e,a\}$, $\displaystyle K_2=\{e,b\}$ and $\displaystyle K_3=\{e,c\}$. By the first isomorphism theorem, $\displaystyle |\mbox{Im}(f)|=\frac{|G|}{|\mbox{Ker}(f)|}$, and therefore whether the kernel of*f*is $\displaystyle K_1$, $\displaystyle K_2$ or $\displaystyle K_3$, the image of*f*must be a subgroup of order 2. Since there are 3 subgroups of*G*of order 2, and three posibilities for the kernel of*f*, there are precisely 9 homomorphisms from*G*to*G*that are non-trivial and not isomorphisms. Since we discovered that there are 6 isomorphisms of*G*earlier (and of course there is one trivial homomorphism of*G*mapping every element of*G*to the identity), there are in total 16 isomorphisms of*G*.

Does this answer your questions? Regarding the idea that a homomorphism is "uniquely determined by its values on the generators of its domain", recall that if*C*is a cyclic group and*f*is a homomorphism on*C*,*f*is determined by its image of the generator of*C*. Why? If*g*generates*C*, and $\displaystyle f(g)=x$ (for some*x*in the range of*f*), $\displaystyle f(g^i)=[f(g)]^i=x^i$. Since every element of*C*is a power of*g*by the very definition of cyclic groups, the image of every element of*C*under*f*is determined by the image of the generator of*C*under*f*. Essentially, this idea generalizes to groups with more than one generator; it might be a good exercise to appreciate this thoroughly. - Feb 25th 2010, 10:53 PMnmatthies1
Thank you! Will go through your posts in more detail over the weekend to see if I fully understand it.

- Mar 31st 2011, 05:07 PMmulaosmanovicben
if an isomorphism maps generators to generators how could f(a)=ab allow f to be an isomorphism?

So far I have this:

Since an endomorphism is uniquely defined by the actions on its generators. And for the klein 4 group the generators are a and b then There are 3 choices for a and 2 choices for b so there are 6 endomorphisms? This does not seem right. Could you guys help me out? - Mar 31st 2011, 07:22 PMDeveno
a homomorphism φ is completely determined by its image on two of the elements of order 2 in G. let's call them a and b, its as good as any other names. so how many ways can we pick a 2-element set out of a 4-element set? 2^4 = 16 ways. 6 of these mappings are automorphisms (that take generators to distinct generators, one of which is the identity map on G).

that map that takes a-->1 b-->1 results in everything going to 1, the trivial map. that leaves 9 other possible homomorphisms.

suppose a-->1. since we have already counted the trivial map, we have 3 choices for the image of b. these map {1,a,b,ab} to {1,φ(b)} with kerφ = {1,a}.

similarly b-->1 yields 3 homomorphisms with {1,a,b,ab} going to {1,φ(a)} with kerφ = {1,b}.

so, what's left? well, the only possibilities are that a and b go to the SAME generator of G, (3 choices here) and we have kerφ = {1,ab}.

that is, all 16 maps of {a,b} to G yield homomorphisms. and these must be the ONLY homomorphisms because φ(1) has to be 1, and φ(ab) has to be φ(a)φ(b). - Apr 1st 2011, 04:37 AMSwlabr
Because $\displaystyle ab$ is a generator! $\displaystyle \langle a, ab\rangle = \langle ab, b\rangle = \langle a, b\rangle$. Indeed, HallsOfIvy's claims that the Klein 4-group has 2 generators is incorrect! It has 3! But it can be generated by 2. That is, you need to work out where a and b are sent, but they can also be sent to ab!

However, I believe you have all the information you need to continue on your own. I mean, in the example Tonio gave you can notice/work out that the function is a homomorphism, and that it is an injection (trivial kernel) and a surjection, and so it is an automorphism!