Let G be a nonabelian group of order 10 prove that G contains an element of order 5 and five element of order 2 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 ?
 A: 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 $p.$
A: 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 $2$.
