# 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 ?

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

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 $p.$
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 $$2$$.