Prove image of symmetric group into additive group of real numbers is zero Suppose $ f : S_{n} \rightarrow (\mathbb{R} , + , 0 , - )$ is a group homomorphism. Prove $ f(S_{n}) = {0} $, i.e., $f(\sigma) = 0$ for every $ \sigma \in S_{n}$with $n \geq 1 $
I cannot seem to find the reason why the transpositions ( $ S_{2} $ ) should have this property. This would of course immediately solve it. Help would be much appreciated!!
 A: Every element of the additive group of real numbers has infinite order except the identity. Elements of finite order, such as all elements in a finite group, can only be mapped to elements of finite order. The result follows.
A: The desired result is actually a specific case of the much more general
Proposition:  Let $G$ be a finite group and $R$ a (not necessarily commutative) unital ring of characteristic $0$ and having no zero divisors; that is, for $r, s \in R$ we have
$rs = 0 \Leftrightarrow r = 0 \;\; \text{or} \;\; s = 0; \tag{1}$
then the only homomophism $f$ from $G$ into the additive group of $R$ is the trivial one, i. e. $f(g) = 0$ for all $g \in G$.
Proof:  The demonstration consists of a few simple steps:
1.)   Let $g \in G$; then $g^m = e$, the identity element of $G$, for some positive integer $m$.  For, $G$ being finite, the elements of the sequence $g$, $g^2$, $g^3$, $\ldots$ of powers of $g$ must at some point begin to repeat themselves; that is, we must have $g^{l + m} = g^l$ for some positive integers $l, m$; but then $g^m = e$.
2.)  For $0 \ne r \in R$ and any positive integer $n$, we have $nr \ne 0$.  Since $\operatorname{char}(R) = 0$, $n = n1_R \ne 0$ in $R$; now if $nr = 0$, we must have $r =  0$ by (1), but this contradicts our hypothesis $r \ne 0$; thus $nr \ne 0$.
3.)  $f(e) = 0$, since $e^2 = e$:
$f(e) = f(e^2) = f(e) + f(e), \tag{2}$
whence
$f(e) = 0 \tag{3}$
as claimed.
4.)  For $g \in G$ and $n$ a positive integer, $f(g^n) = nf(g) \in R$; a simple induction establishes this, if $f(g^k) = kf(g)$, then 
$f(g^{k + 1}) = f(g^k g) = f(g^k) + f(g)$
$= kf(g) + f(g) = (k + 1)f(g); \tag{4}$
we take $f(g) = f(g^1) = 1f(g) = f(g)$ as the base case.  
5.)  Now suppose $f(g) = r \ne 0$; since by item (1.), $g^m = e$ for some $m \ge 1$, by items (3.) and (4.) we have
$mf(g) = f(g^m) = f(e) = 0, \tag{5}$
in contradiction to item (2.); therefore,
$f(g) = 0, \tag{6}$
that is, $f:G \to (R, +, 0, -)$ is trivial.  QED.
The precise case at hand is now resolved by taking $G = S_n$ and $R = \Bbb R$.
