1
$\begingroup$

Because order of element divide order of the group,the possiblility of order of element is 1,2,5,10. Frin this ,it must contains element of order 5 and 2. But I can't prove that it must gaurantee five elements of order 2 .Anyone can give a hint ?

$\endgroup$
1
  • 2
    $\begingroup$ Hint: How many Sylow 5 and 2 subgroups it must have? $\endgroup$ – DonAntonio Dec 14 '16 at 8:54
2
$\begingroup$

Hint: Look at Sylow $2$ and $5$ subgroups.

EDIT: Also notice that the statement of your problem follows trivially from Cauchy's theorem which says that if $p$ is a prime and $p$ divides the order of a group, then the group contains an element of order $p.$

$\endgroup$
1
$\begingroup$

By Lagrange, the possible orders of the non-identity elements of the non-abelian $G$ are $2$ and $5$. Not all the nonidentity elements have order $2$, otherwise, $G$ would be abelian. So $G$ must contain an element "$a$" that has order $5$. Therefore the elements of $G$ that are of order $5$ are $a, a^2, a^3, a^4$. To eliminate the possibility of there being any others suppose that there exists $b \in G$ such that b does not belong to $\langle a \rangle$ and $|b|=5$ since $5$ is prime no power of $b$ will belong to $\langle a \rangle$. Now consider the two cosets $b^2 \langle a \rangle$ and $b^3 \langle a \rangle $. Since $|\langle a \rangle|=5$ there are only two distinct cosets one of which is $\langle a \rangle$. Therefore $b^2 \langle a \rangle = b^3\langle a \rangle$ which implies that $b^2b^{-3} \langle a \rangle = \langle a \rangle$ implying $b^2b^2 \langle a \rangle = \langle a \rangle$ implying that $b^4 \in \langle a \rangle$ contradiction. The conclusion is that $4$ elements have order $5$ and $5$ elements have order $2$.

$\endgroup$
1
  • $\begingroup$ This did not copy right $\endgroup$ – Greg Jan 28 '20 at 3:14

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.