# Homomorphism between free group and symmetric group

Let $G:=S_5$ with elements $g_1=(12345)$ and $g_2=(12)$ and let $\varphi:F_2\to G$ be the unique homomorphism satisfying $\varphi(x_i)=g_i$ for $i=1,2$

Questions

$\bullet$ Is $\varphi(x_2^{-1})=\varphi(x_2)^{-1}$ here ?

$\bullet$ How can I find an element $w$ of $F_2$ such that $\varphi(w)=(34)$

In general $g_1$ and $g_2$ generate $G$, to write $(34)$ in terms of $g_1,g_2$, I did:

$\varphi(w)=(34)=(14)(13)(14)=\underbrace{\Big((13)(12345)(13)\Big)}_{(14)}(13)\Big((13)(12345)(13)\Big)$

$=(13)(12345)(13)(12345)(13)\tag I$

$$=\underbrace{\Big((12)(12345)(12)\Big)(12)\Big((12)(12345)(12)\Big)^{-1}}_{(13)}(12345)...$$

but $((12)(12345)(12))^{-1}=(12)^{-1}(12345)^{-1}(12)^{-1}=(12)(12345)^4(12)$

and so if $g_1$ corresponds to $x_1$, then $(13)$ corresponds to $x_2x_1x_2x_1^4x_2$ and the word $w$ is just the ''juxtaposition'' of these $2$ according to $(I)$, is that correct ?

• Any homomorphism fulfills your first question, so why to ask in this case? As for the 2nd. question: find a representation of $\;(3\;4)\;$ as a product of $\;(1\;2)\;,\;\;(1\;2\;3\;4\;5)\;$ and pull back to $\;F_2\;$, so yes: you're correct (I didn't check to depth your calculations) – Timbuc Nov 16 '14 at 13:05

First question: Yes, because $\phi$ is a homomoprhism.
Second: Yes, simply put if you find an expression in the $g_i$ that produces a desired element of $G$, just replace all letters "$g$" with "$x$" and you obtain an element of $F_2$ that is mapped as desired. (Actually, I would have found a different solution by conjugating $g_2$ with $g_1^2$, but of course there are many possible solutions ...)