Question on Lagrange theorem

Question

Find order of $H$ if $H$ is subgroup of some group of order $100$ and $H$ contains no element of order $2$, with $H$ non cyclic

The ans given in a book is $25$. I can understand that $\operatorname{O(H)}$ divides $\operatorname{O(G)}$

I couldn't understand the below point: since $H$ is non-cyclic, therefore $\operatorname{O(H)}$ is not equal to $1,2$ and $5$ so the possibilites for $\operatorname{O(H)}$ are $4,10,20,25,50 \text{ and } 100$ Now ,groups of even order must contain an element of order $2$ since $H$ does not ,it must have odd order thus $\operatorname{O(H)} =25$

Answer 1

The solution is correct. The existence of an element of order $2$ would follow from Cauchy's theorem, in case $H$ had even order. That enables you to close it out.

Note that, essentially by Lagrange, groups of prime order are cyclic.

Answer 2

Solution seems to be ok. $H$ cannot be of prime order since Group of prime order is cyclic. Also note that If G is a group of even order then number of elements of order two is odd(exercise) . Given order of group is $100$. Divisors are possibilities for subgroup not exactly subgroup. If order of $H$ is even note that Cauchy's theorem states that if $G$ is a finite group and $p$ is a prime number dividing the order of $G$ (the number of elements in $G$), then $G$ contains an element of order $p$ . So the only possible order of $H$ is $25$ which divides $100$