Let $G$ be an finite abelian group, define $s= \Pi_{g\in G} g$, show that $s = e$ unless $G$ has exactly one involution $u$, in this case $s=u$.   I have the start of a proof, but I get stuck at a point.   Proof  Write $G$ as $G = \{e\} \cup I \cup \{g_1, g_1^{-1}, \ldots, g_n, g_n^{-1}\}$ where $I$ is a set of the all the groups involutions.  Since $G$ is abelian $\Pi_{g\in \{g_1, g_1^{-1}, \ldots\}} g = e$ we just need to examine $\Pi_{u\in I} u$  Now:   If $|I|=0$ then the theorem holds, If $|I| = 1$ then $I=\{u\}$ and $s= u$, the theorem holds If $|I| = 2$, requires closer examination:  Let $I = \{u_1,u_2\}$ be the set of all involutions (not equal to $e$), then $u_1u_2$ is also an involution (since $G$ is abelian).   If $u_1u_2 =e$ then $u_1=u_2$, contradiction, if $u_1u_2 = u_1$ then $u_2=e$, contradiction. $u_1 u_2=u_2$ results in the same contradiction. This means that the case $|I|=2$ is not possible. the case $|I|=3$ is possible, let $I = \{u_1,u_2,u_3\}$ then it's easy to prove $u_3 = u_1u_2$. The theorem holds.   How should I continue? I've tried splitting $I$ as   $$I=\{u_i: u_i \not = u_ju_k (\text{for a certain }j\not = k)\} \cup \{u_i = u_ju_k (\text{for a certain} j \not = k)\}$$ but I still get stuck.  Also the case $|I|= 4$ seems contradictory once again..
Suppose $G=\{a_1,...,a_n\}$ is a finite abelian group, and let $x=a_1a_2\dotsm a_n$.   Prove that if there is more than one element of order $2$ then $x=e$.   What I've done so far: (#1 is just for illustration, There should be at least $3$ such elements)   $G$ has an even number of elements with order $&gt;2$ (They are paired so: $(a,a^{-1})$ ). And then there are $e$ and $\{b|o(b)=2\}$. Also, if $G$ has at least one element of order 2 then $|G|$ is even (Lagrange). Thus $|\{b|o(b)=2\}|$ must be odd. If there is exactly one element $b$ of order $2$ then $x=b$. That's because $G$ is abelian, and we can write: $x=e\cdot b\cdot(a_1 a_{1}^{-1}\dotsm a_n a_{n}^{-1})=b\cdot(e\dotsm e)=b$. If there are exactly $3$ such elements $a,b,c\in G$, then, as shown above, $x=abc$ and thus $x=e$ (That's because $(abc)^2=a^2b^2c^2=e$ which means $abc\in \{a,b,c,e\}$. If $abc=a$ then $b=c^{-1}=c$ which is false, same goes for $abc=b,abc=c$.)   I haven't managed to find a reason why the claim must hold for $5, 7, ...$ (i.e. All other odd integers). Does anyone have an idea?  Thanks in advance. Please excuse my English.  edit: Not a canonical answer but a simple one (if such even exists).
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