# What is the kernel of $h$ where $h:S_3 to mathbb{R}^*$

$$S_3$$ is the permutation group on 3 symbols and $$mathbb{R}^*$$ denotes the multiplicative group of non zero real numbers under multiplication. $$h: S_3 to mathbb{R}^*$$is given a homomorphism. But I am unable to define the mapping. I don’t know which mapping will make it a homomorphism. Only then I’ll be able to find the kernel. What am I missing?

Mathematics Asked by user686123 on November 21, 2021

If $$h: S_3 to R^*$$ is a homomorphism, then $$S_3/textrm{kern}(h) cong h(S^3) subseteq R^*$$. For $$sigma, tau in S^3$$, we would then have $$h(sigma tau sigma^{-1} tau^{-1}) = h(sigma) h(tau) h(sigma^{-1}) h( tau^{-1}) = h(sigma) h(tau) h(sigma)^{-1} h( tau)^{-1} = 1$$ by the commutativity of $$R^*$$. Therefore $$textrm{kern(h)}$$ contains the commutator subgroup of $$S_3$$. This subgroup happens to be the cyclic group generated by $$(1;2;3)$$. By lagrange's theorem, since the order of $$textrm{kern(h)}$$ is at least 3, we have that $$|h(S_3)| leq 2$$ which leaves few options for $$h$$.

Answered by Daniel on November 21, 2021

The imagine of $$S_3$$ must be a subgroup of $$Bbb R ^*$$ of order a divisor of 6. So, if $$xin h(S_3)$$ $$x^6=1implies x=pm 1$$. Now we have two possibilities: $$h(S_3)={1}$$ or $$h(S_3)={pm 1}$$. In the first case the kernel is all the group, in the second it must be a subgroup of $$S_3$$ of index 2, so it must be $$A_3$$.

Answered by Alessandro Cigna on November 21, 2021

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