# Thread: Number of different homomorphisms

1. ## Number of different homomorphisms

How many homomorphisms exist from the abelian group $Z_2\oplus Z_2$ to $S_3$.

Since for every $1\neq a\in Z_2\oplus Z_2$ we have $a^2=1$ then we must have $f^2(a)=1$ ( $f$ is an homomorphism). so $f(a)$ must be $(12)$ or $(23)$ or $(13)$ but since a product of any of the above is not of order 2 then $f$ would not be a homomorphism.

Does it seem right or am I fundamentaly wrong?

Any help appreciated

2. Originally Posted by skyking
How many homomorphisms exist from the abelian group $Z_2\oplus Z_2$ to $S_3$.

Since for every $1\neq a\in Z_2\oplus Z_2$ we have $a^2=1$ then we must have $f^2(a)=1$ ( $f$ is an homomorphism). so $f(a)$ must be $(12)$ or $(23)$ or $(13)$ but since a product of any of the above is not of order 2 then $f$ would not be a homomorphism.

Does it seem right or am I fundamentaly wrong?

Any help appreciated

You are wrong but I think not fundamentally: your argument about the order of the

image is right, but you oversaw the fact that the order of any of the three 2-cycles

you wrote there indeed is 2, so you can define $f(a)=(12)\,\,or\,\,(13)\,\,or\,\,(23)$, and

the same can be done with any of the other two elements of order 2 in that direct sum.

Of course, the above is not the whole story since you still have

to figure out how to define f on the other elements of that sum...

Tonio

3. Also, a "trivial" point: the zero homomorphism exists between any two groups.

4. Originally Posted by tonio
You are wrong but I think not fundamentally: your argument about the order of the

image is right, but you oversaw the fact that the order of any of the three 2-cycles

you wrote there indeed is 2, so you can define $f(a)=(12)\,\,or\,\,(13)\,\,or\,\,(23)$, and

the same can be done with any of the other two elements of order 2 in that direct sum.

Of course, the above is not the whole story since you still have

to figure out how to define f on the other elements of that sum...

Tonio
OK, so this is what I get. Define $Z_2\oplus Z_2 =\{(0,0),(0,1),(1,0),(1,1)\} =\{ e,a,b,c\}$, so in order for $f$ to be a homomorphism we must define $f(e)=(1), f(a)=(mn), f(b)=(jk), f(c)=(rs)$ (where $(mn),(jk),(rs)$ are 2-cycles in $S_3$). But the we get $f(a+b)=f(c)\neq f(a)f(b)$ since $(mn)(jk)$ is not a 2-cycle. similarly we can show the same for other combinations of $a,b,c$. Therefore $f$ is not a homomorphism.

So concluding we get that the only homomorphism is the trivial one.

Am I missing something here?

5. What happens if you send a single nonidentity element of Z2+Z2 to a 2-cycle, and send everything else to the identity?

6. Originally Posted by Tinyboss
What happens if you send a single nonidentity element of Z2+Z2 to a 2-cycle, and send everything else to the identity?
It still would not be a homomorphism. for example if we define $f(a)=(12)$ and everything else goes to $(1)$, then we get $(1)=f(c)=f(a+b)\neq (12)(1)=f(a)f(b)$.

But if we choose 2 nonidentity elemnts from $Z_2\oplus Z_2$ and send them to the same 2-cycle (for example $(12)$) then we will get a homomorphism. For instance $f(a)=f(b)=(12)$, then $(1)=f(c)=f(a+b)=f(a)f(b)=(12)(12)$ and $(12)=f(b)=f(a+c)=f(a)f(c)=(12)(1)$ and similarly for other combinations.

So now in addition to the trivial homomorphism we get $3\times 3$ choices for different homomorphisms.

I think I am not missing anything now.

Thanks