My idea was to use the homomorphism theorem for [$S_n$,$S_n$] as for any f:$S_n$ → C* by setting N=[$S_n$,$S_n$] we know that there is exacly one g:$S_n$/[$S_n$,$S_n$] → C*  However we know that $S_n$/[$S_n$,$S_n$]=$S_n$/$A_n$, but $S_n$/$A_n$ ≅ Z/2Z  Therefore there is exacly one g:Z/2Z → C* so all elements of a ∈ C* give us ord(a)|g[Z/2Z] so a can only be -1 or 1.   Now we know that f:$S_n$ → {-1,1}  Aaaaand I'm stuck. :/
I need to find all group homomorphisms of $S_n\to\mathbb{C}^*$ (for $n\ge2$), and also all group homomorphisms of $A_n\to\mathbb{C}^*$ (for $n\ge2$).  Someone on this site, two years ago, seems to have already asked the same question (). I understand what the questioner says there, but I don't understand it when he says:     Therefore there is exactly one $g:\mathbb{Z}/2\mathbb{Z}\to\mathbb{C}^*$, so all elements of $a\in\mathbb{C}^*$ give us $\operatorname{ord}(a)\mid g[\mathbb{Z}/2\mathbb{Z}]$, so $a$ can only be $-1$ or $1$. Now we know that $f:S_n\to\{-1,1\}$ &hellip;   What is he saying there? And how to go on from there? (I do not understand the answer given either &hellip;)  I hope that someone here can help me further. (And if I'm not allowed to ask this question because is has already been asked two years ago, please tell me what to do instead).  Thank you in advance!
Convergence of sequence of real numbers and (sub)subsequence Let $(x_n)_{n\in \mathbb{N}}$ be a sequence of real numbers. A well known result is:     The sequence $(x_n)_{n\in \mathbb{N}}$ converges to $a\in \mathbb{R}$ if and only if every subsequence $(x_{n_k})_{k\in \mathbb{N}}$ converges to $a\in \mathbb{R}$.   Now, I'm working on an exercise which essentialy asks to prove the following:     The sequence $(x_n)_{n\in \mathbb{N}}$ converges to $a\in \mathbb{R}$ if and only if for every subsequence $(x_{n_k})_{k\in \mathbb{N}}$ there is a (sub)subsequence $(x_{n_{k_j}})_{j\in \mathbb{N}}$ which converges to $a$.   Now, one direction is trivial. If in fact $(x_n)_{n\in \mathbb{N}}$ converges to $a$, then by the result I've mentioned, every subsequence of $(x_{n})_{n\in \mathbb{N}}$ converges to $a$. Let then $(x_{n_k})_{k\in \mathbb{N}}$ be an arbitrary subsequence. In that case, we know that it is a subsequence of itself, and hence we've found a (sub)subsequence which converges to $a$.  The other way around seems quite complicated for me. The reason is as follows: I must assume that every subsequence $(x_{n_k})_{k\in \mathbb{N}}$ has a (sub)subsequence which converges to $a$ and the proof that the original one converges to $a$.  My problem is exactly that "a". If it had all (sub)subsequences converging to $a$, then by the result i've mentionted, the subsequence $(x_{n_k})_{k\in \mathbb{N}}$ would converge to $a$. In that case, we would have every subsequence converging to $a$ and the result would follow.  My idea to prove the claim was then:  First remember that a subsequence is the restriction of the original one $x: \mathbb{N}\to \mathbb{R}$ to an infinite subset $\mathbb{N}'\subset \mathbb{N}$. In that case a (sub)subsequence is the restriction of $\tilde{x} = x|\mathbb{N}'$ to an infinite subset $\mathbb{N}''\subset \mathbb{N}$. 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