# Be $G=\{ e,g_1,g_2,\ldots, g_n \}$, $|G|=n+1$. Suppose $G$ has a unique element of order $2$, say $g_1$. Show that $eg_1g_2\ldots g_n=g_1$.

Be $G=\{ e,g_1,g_2,\ldots, g_n \}$ an abelian group of order $n+1$. Suppose $G$ has a unique element of order $2$, say $g_1$. Show that $eg_1g_2\ldots g_n=g_1$.

I have serious difficulties with algebra, I'm trying to study, but would help to understand and believe that an exercise is a great way as well.

If $g_1$ has order $2$, $g_2^2=e$, i.e. $g$ is self-inverse, so $g_1^{-1}=g_1$.
No other elements $g_2,g_3,...,g_n$ are self inverse as $g_1$ is the unique self-inverse element. So in the product $$eg_1g_2\cdots g_n$$ , rearrange $g_2g_3\cdots g_n$ so that each of the $g_2,g_3,...,g_n$ is next to its inverse (we can do this as the group is abelian). $g_1$ is self-inverse so can't be paired off with an element in $g_2, g_3, ...,g_n$. Thus $g_2g_3\cdots g_n=e$, and $eg_1(g_2g_3 \cdots g_n)=eg_1 e= g_1$.
Note that as every element in $S=\{g_2,g_3,...,g_n\}$ can be split into pairs, $|S|$, and so $|G|=|S \cup{e,g_1}|$ is even.
• Thank you very much. I tried to do here again, only considering $n$ is odd (the way you did), and then $n$ is even, so finding an absurdity. Continues correct? – marcelolpjunior Apr 23 '14 at 19:26
• @marcelolpjunior You mean you proved it for $n$ odd, then discarded the case of $n$ even because $n$ is never even? If so, that's correct. – Meow Apr 23 '14 at 19:27