# Homomorphisms

• May 19th 2008, 01:25 PM
alittletouched
Homomorphisms
Hello all. I am stuck trying to figure out if these are homomorphisms and what there images/kernels are if that is the case. Any help would be great! James

$G=\Sigma_5, X=\Sigma_5$. Then the homomorphism takes $\sigma\rightarrow (3 5 2 4 1)*\sigma*(3 1 4 2 5)$. I know that this is a homomorphism, I am just stuck on the kernel/image

Working in $\Sigma_8$. Then $G=<(1 2 3)(8 4 6 7 5)>$. If $\sigma=(1 2 3)(8 4 6 7 5)$, the homomorphism maps to $\mathbb{Z}_5$ by $\sigma^i\rightarrow i+5\mathbb{Z}$. I'm not sure what to do from here

Thanks
• May 19th 2008, 02:53 PM
Jhevon
Quote:

Originally Posted by alittletouched
Hello all. I am stuck trying to figure out if these are homomorphisms and what there images/kernels are if that is the case. Any help would be great! James

$G=\Sigma_5, X=\Sigma_5$. Then the homomorphism takes $\sigma\rightarrow (3 5 2 4 1)*\sigma*(3 1 4 2 5)$. I know that this is a homomorphism, I am just stuck on the kernel/image

Working in $\Sigma_8$. Then $G=<(1 2 3)(8 4 6 7 5)>$. If $\sigma=(1 2 3)(8 4 6 7 5)$, the homomorphism maps to $\mathbb{Z}_5$ by $\sigma^i\rightarrow i+5\mathbb{Z}$. I'm not sure what to do from here

Thanks

How is $\Sigma_n$ defined?
• May 19th 2008, 03:54 PM
alittletouched
$\Sigma_n$ is the symmetric group of n letters

As examples

$\Sigma_2=\left\{1_{\Sigma_2},(1 2)\right\}$
and
$\Sigma_3=\left\{1_{\Sigma_3},(1 2),(2 3),(1 3),(1 2 3),(1 3 2)\right\}$

Thank! James
• May 19th 2008, 04:11 PM
Jhevon
Quote:

Originally Posted by alittletouched
Hello all. I am stuck trying to figure out if these are homomorphisms and what there images/kernels are if that is the case. Any help would be great! James

$G=\Sigma_5, X=\Sigma_5$. Then the homomorphism takes $\sigma\rightarrow (3 5 2 4 1)*\sigma*(3 1 4 2 5)$. I know that this is a homomorphism, I am just stuck on the kernel/image

Oh! i always used the notation $S_n$. anyway, the Kernel of the homomorphism is only contains the identity.

note that $(3~5~2~4~1)^{-1} = (3~1~4~2~5)$

to get an output of $(1)$ we need $\sigma * (3~1~4~2~5) = (3~5~2~4~1)^{-1}$

clearly $\sigma$ has to be the identity permutation
• May 19th 2008, 04:20 PM
Jhevon
Quote:

Originally Posted by alittletouched
Working in $\Sigma_8$. Then $G=<(1 2 3)(8 4 6 7 5)>$. If $\sigma=(1 2 3)(8 4 6 7 5)$, the homomorphism maps to $\mathbb{Z}_5$ by $\sigma^i\rightarrow i+5\mathbb{Z}$. I'm not sure what to do from here

Thanks

as for this one. if we are mapping to $\mathbb{Z}_5$ then we should be mapping to equivalence classes mod 5, right?

so you mean $\sigma^i \mapsto [i + 5 \mathbb{Z}]$

the order of $\sigma$ is 8, so that $0 \le i \le 8$, and then everything repeats for higher values. thus, to find the kernel (which i suppose is what you want), it would be $\{ \sigma^i \mid i + 5 \mathbb{Z} \text{ is a multiple of 5}\}$
• May 19th 2008, 08:51 PM
NonCommAlg
Quote:

Originally Posted by Jhevon
as for this one. if we are mapping to $\mathbb{Z}_5$ then we should be mapping to equivalence classes mod 5, right?

so you mean $\sigma^i \mapsto [i + 5 \mathbb{Z}]$

the order of $\sigma$ is 8, so that $0 \le i \le 8$, and then everything repeats for higher values. thus, to find the kernel (which i suppose is what you want), it would be $\{ \sigma^i \mid i + 5 \mathbb{Z} \text{ is a multiple of 5}\}$

just a couple of points:

1) the order of $\sigma$ is 15 not 8.

2) it is important to note that $\sigma^i \rightarrow [i + 5\mathbb{Z}]$ is basically well-defined. that's because 15 (the order of $\sigma$) is divisible by 5.

3) then the map is obviously homomorphism and onto and its kernel is $\{(1), (1 \ 2 \ 3), (1 \ 3 \ 2) \}.$
• May 19th 2008, 08:54 PM
Jhevon
Quote:

Originally Posted by NonCommAlg
just a couple of points:

1) the order of $\sigma$ is 15 not 8.

yes, you're right. i added when i should have multiplied :p
• May 20th 2008, 01:26 AM
alittletouched
That's great thanks :)

How would I find the images for the homomorphisms? I expect that the image of the second homomorphism is merely the integers but Im not sure about the first (Lipssealed)