1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$ – marcelolpjunior Apr 23 '14 at 19:26
  • $\begingroup$ @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. $\endgroup$ – Meow Apr 23 '14 at 19:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.