# How to find kernel of this map: is my attempt correct?

Find $\ker(\varphi)$ and $\varphi(3,10)$ for $\varphi \colon \mathbb Z\times\mathbb Z\to S_{10}$ such that $\varphi(1,0)=(3,5)(2,4)$ and $\varphi(0,1)=(1,7)(6,10,8,9)$. For example if I get order of $\varphi(1,0)$ is maybe $3$ and that of $\varphi(0,1)$ is maybe $4$, then should I write $(3,4) (\mathbb Z \times \mathbb Z)$ as the kernel of $\varphi$?

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The order of $\phi(1,0)$ is $2$ as it is the product of two disjoint cycles of length two, while the order of $\phi(0,1)$ is $4$ as it is the product of two disjoint cycles, the lengths of which have lcm $4$. This means $\operatorname{ker}(\phi) = (2,4)(\mathbb{Z}\times \mathbb{Z})$, as if $\phi(a,b) = (1)$ then $\phi(1,0)^a\phi(0,1)^b = (1)$ since the cycles $\phi(1,0),\phi(0,1)$ are disjoint hence commute, and because the cycles $\phi(1,0)^a\phi(0,1)^b$ are disjoint they must both be $(1)$.
To compute $\phi(3,10)$, note that $\phi$ is a homomorphism so $$\phi(3,10) = \phi(1,0)\phi(1,0)\phi(1,0)\phi(0,1)\cdots\phi(0,1) = \phi(1,0)^3\phi(0,1)^{10}$$ and because these have order two and four respectively this is equivalent to $\phi(1,0)\phi(0,1)^2$. This should be easy enough to compute by hand.
A crucial feature of $\varphi$ for the argument above to be valid is that $\varphi(1,0)$ and $\varphi(0,1)$ have disjoint supports. Exercise: enumerate the steps which become false if $\varphi(1,0)=(1,2)$ and $\varphi(0,1)=(2,3)$. – Did Dec 25 '11 at 23:51
Nitpick: The order of $\phi(0,1)$ is 4 because it is the product of disjoint cycles whose lengths have lcm 4, not just because the longest cycle in the product has length 4. – Alex Dec 26 '11 at 2:01
@Alex I was just about to say that. As for the question on notation, personally I would write the kernel as $2\mathbb{Z}\times 4\mathbb{Z}$, or maybe just $(2, 4)$, but your notation is perfectly acceptable. It's not really misunderstandable, and avoiding misunderstandings is the main reason for a strict notation system. – Arthur Dec 26 '11 at 2:10