7
$\begingroup$

Suppose there exists such a group. Then Lagrange's theorem assures that the group is of even order.

But I conclude from this and this that such a group has odd number of elements of order $2$. Giving us contradiction.

Hence there does not exist a finite abelian group $G$ containing exactly $60$ elements of order $2$.

More strongly there does not exist a finite group $G$ containing even number of elements of order $2$.

Is my understanding correct?

$\endgroup$
  • $\begingroup$ Did you get this question from TIFR entrance ?? $\endgroup$ – Anik Bhowmick Oct 8 '18 at 3:56
  • $\begingroup$ Yes GS 2017 . {}{}{}{}{} $\endgroup$ – blue boy Dec 7 '18 at 6:00
6
$\begingroup$

Yes, your understanding is correct.

Consider the relation $\sim$ on $G$ (having even order, otherwise it has no element of order $2$) defined by $$ a\sim b \quad\text{if and only if}\quad (b=a\text{ or }b=a^{-1}) $$ This is easily seen to be an equivalence relation. The equivalence classes have either one or two elements. If you remove the two-element equivalence classes, you are dropping an even number of elements from $G$, so what remains is an even number. Drop also the class consisting of $1$ and you remain with an odd number of one-element equivalence classes: these elements are precisely those having order $2$.

$\endgroup$

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.