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

Aug 2013
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!


Country Boy

Math Team
Jan 2015
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
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:
$h(2,0)=(1,2)^2=e\text{, the identity of }S_3$
$h(1,1)=(1,2)(1,2,3)=(2,3)$ (composition from right to left)

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,

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