Proving or disproving homomorphism from that map that maps (Z x Z) to S_3?

Aug 2013
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0
In mathematics, given two groups, (G, ∗) and (H, ·), a group homomorphism from (G, ∗) to (H, ·) is a function h : G → H such that for all u and v in G it holds that h(u*v)=h(u)h(v). I have never done a mapping onto a permutation group though, so I am not sure how to approach this problem. Thank you in advance!
 

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Country Boy

Math Team
Jan 2015
3,261
899
Alabama
The function here maps the pair of integers (i, j) to the permutation \(\displaystyle (12)^i(123)^j\). I presume that the operation on ZxZ is (i,j)*(m,n)= (i+m, j+ n). You want to show that, for any (i, j) and (m, n) in ZxZ, \(\displaystyle [(12)^i(123)^j][(12)^m(123)^n]= (12)^{i+m)(123)^{i+m}\).

Now, do you know what "(12)" and "(123)" mean here?
 
Jan 2016
93
48
Athens, OH
Give the name $h$ to the mapping; i.e. $h(i,j)=(1,2)^i(1,2,3)^j$. The function h is not a homomorphism. First h is a surjective map:
$h(1,0)=(1,2)$
$h(0,1)=(1,2,3)$
$h(2,0)=(1,2)^2=e\text{, the identity of }S_3$
$h(1,1)=(1,2)(1,2,3)=(2,3)$ (composition from right to left)
$h(0,2)=(1,2,3)^2=(1,3,2)$
$h(1,2)=(1,2)(1,2,3)^2=(1,2)(1,3,2)=(1,3)$

The above equations show that every element of $S_3$ is an image.

Fact: if h is a homomorphism from an abelian group G onto a group H, then H is abelian. You can try and prove this. So, if h were a homomorphism, then $S_3$ would be abelian, but it's not.

Alternatively, if h were a homomorphism,
$h(1,1)=h((1,0)+(0,1))=h(1,0)h(0,1)=(1,2)(1,2,3)$
$h(1,1)=h((0,1)+(1,0))=h(0,1)h(1,0)=(1,2,3)(1,2)$

But $(1,2)(1,2,3)\neq((1,2,3)(1,2)$
 
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