the number of the subgroups of a non cyclic group whose order is $25$ My question is about group theory:

How many subgroups does a non-cyclic group contain whose order is 25?

How can i answer that question?
Can you generalize the answer?
Thanks for your help.
 A: As a power of a prime, a non-cyclic group $\;G\;$ of order $\;25\;$ is isomorphic to $\,C_5\times C_5\;$ , with $\,C_5\cong\Bbb Z/5\Bbb Z\;$ is the cyclic group of order $\;5\;$
Now, some hints for you to answer your question:
== A group of the form $\;G=\underbrace{C_p\times C_p\times\ldots\times C_p}_{n\;\text{times}}\;$ is a vector space of dimension $\;n\;$ over the field $\,\Bbb Z/p\Bbb Z\;$ and is therefore isomorphic to $\;\left(\Bbb Z/p\Bbb Z\right)^n\;$
== The subgroups of $\,G\,$ as above are in $\,1-1\,$ correspondence with the vector subspaces of $\,\left(\Bbb Z/p\Bbb Z\right)^n\;$
A: There are three possibilities for the order of any subgroup $H$ of a group $G$ of order $25$:


*

*$|H| = 1 \iff H = e$

*$|H| = 5,$ since $5\mid 25$.

*$|H| = 25$, if $H = G$.


We're given that $G$ is non-cyclic, so the order of any element $x \neq e$ must be $5$, else, if $25$, it would generate the group, and hence the group would be cyclic. (Contradiction). I.e., $x \neq e \;  \implies \;|\langle x \rangle| = 5$, and for each distinct subgroup $\langle x_i\rangle = \{e, x_i, x_i^2, x_i^3, x_i^4\}$, the elements $x_i, x_i^2, x_i^3, x_i^4$ are each of order $5$, and any one of them generates the same subgroup as does $x_i$.
Since $G$ is non cyclic, and $|G| = 25 = 5^2,$ where $5$ is prime, we know that $G \cong \mathbb Z_5 \times \mathbb Z_5$
A: If $H$ is a non-trivial subgroup, then $H$ has $5$ elements (why?).
As the subgroup is not cyclic, and $25=5^2$, the order of every element $x \neq e$ is .....
Last but not least, if the order of $x$ is 5, there are 4 other elements of order 5 which generate the same group.
