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Problem Statement: Prove that the one-dimensional characters of a group $G$ form a group under multiplication of functions, i.e. where the group operation is: $$(\chi\cdot\chi')(g)=\chi(g)\chi'(g)$$ This group is called the character group of $G$, and is often denoted by $\hat{G}$. Prove that if $G$ is abelian, then $|\hat{G}|=|G|$.

So, I completely understand how to prove that $\hat{G}$ is a group, but I am unsure about the second part: proving that if $G$ is abelian, then $|\hat{G}|=|G|$.

Clearly, $\hat{G}$ is abelian regardless of $G$ being abelian, since $\hat{G}\subset\mathbb{C}^\times$ and $\mathbb{C}^\times$ is commuative. So $G$ being abelian eliminates the possibility that $(\chi\cdot\chi')(gh)=(\chi\cdot\chi')(hg)$ with $gh\neq hg$.

So basically, to prove that $|\hat{G}|=|G|$, we must show that for any distinct $g,h\in G$, then $(\chi\cdot\chi')(g)\neq(\chi\cdot\chi')(h)$, but I am unsure how I should approach this type of proof.

I was thinking somehow prove that there is an isomorphism between $G$ and $\hat{G}$, but my professor said that constructing a homomorphism between the two groups would be too difficult without using material that we haven't learned yet.

Any suggestions for showing that $|\hat{G}|=|G|$ are appreciated!

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    $\begingroup$ Have you studied the structure theorem for finitely generated abelian groups? If you can show it is true for cyclic groups, and show that $|\widehat{G\times H}| = |\widehat G| \times |\widehat H|$ for groups $G,H$, then the result follows. $\endgroup$ – Mathmo123 Feb 14 '16 at 21:15
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    $\begingroup$ No, I don't believe we have studied that. I am not familiar with that notation. One thing also confusing me, one-dimensional characters are defined by the group $G$, since the elements in a representation of $G$ are just $1\times 1$ matrices so for $A\in G$, $\chi(A)=\mathrm{tr}(A)$ which is just the matrix entry. Wouldn't that mean that $G$ is always abelian? $\endgroup$ – yung_Pabs Feb 14 '16 at 21:16
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    $\begingroup$ The notation is just your notation (my hats are wider than yours) - $\widehat{ G\times H}$ is the group of characters of the group $G\times H$. The structure theorem states that every finitely generated abelian group is a product of cyclic groups (with some extra properties). I guess this is what your professor was referring to by material you haven't learnt yet. $\endgroup$ – Mathmo123 Feb 14 '16 at 21:17
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    $\begingroup$ See also Proposition $2.2.3$ of Bump's lecture notes cited here. $\endgroup$ – Dietrich Burde Feb 14 '16 at 21:17
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    $\begingroup$ In answer to your second question, the term character has two meanings: first is a group homomorphism from $G\to \mathbb C^\times$. The second is the trace of a group representation. The latter is not generally a group homomorphism, since $\mathrm{tr}(AB)$ is not generally equal to $\mathrm{tr}(BA)$. $\endgroup$ – Mathmo123 Feb 14 '16 at 21:21

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