Let G be group of order $125$. Will it have a subgroup of order $25$? Let $G$ be a group of order $125$. Does it have a subgroup of order $25$ ? For any abelian group, the answer is yes. But what about non-abelian groups? I cannot understand. Can anyone help me?
 A: Sylow's First Theorem: Let $G$ be a finite group and $p$ be a prime. Then for each $p^k$ dividing $|G|$, $G$ has a subgroup of order $p^k$.
From this it directly follows that, $G$ has subgroups of order $5$ and $5^2=25$.
I would also want you to note the following observations and prove the following results as a handy exercise.
Result-1: The center of a Group is never a maximal subgroup.
Result-2: If $|G|=p^k$, where $p$ is prime then $G$ has a non-trivial center. 
From the above results, we can go on to say that $|Z(G)|=5$.
A: The “big property” you need is that a group of order $p^n$ ($p$ a prime, $n>0$) has a non trivial center. This is one of the most important facts about $p$-groups.
The proof of the fact is a consequence of the conjugacy class equation.
Once you accept it, it's quite simple to prove the following proposition. 

Let $G$ be a $p$-group of order $p^n$. If $0\le k<n$, then there exists a subgroup of $G$ of order $p^k$.

Proof. The case $n=1$ is obvious. So let's assume the statement holds for every $p$-group of order $p^{n-1}$, with $n>1$.
By the fact mentioned above, $|Z(G)|=p^z$, with $z>0$, so it has a subgroup $Z_0$ of order $p$ which is obviously normal in $G$.
There is nothing to prove for $k=0$, so assume $0<k<n$. Since $G/Z_0$ has order $p^{n-1}$, by induction hypothesis it has a subgroup $H/Z_0$ of order $p^{k-1}$. Therefore $|H|=p^k$. QED
A: Groups with order a power of a prime (i.e. $p$-groups) have the 

Property: have a normal subgroup of order $d$ for every $d$ that divides the order of the group. 

This is not a coincidence, indeed a finite $p$-group is nilpotent and finite nilpotent groups are characterized by the above property (remember that finite nilpotent groups are direct product of their $p$-Sylow subgroups). This is no longer true if we relax nilpotency to solvability ($S_3$ with $p=2$).
If we do not care about normality, we may ask whether a group $G$ has a subgroup of order $d$ for every $d$ that divides $|G|$. This called CLT property (i.e. Converse of the Lagrange Theorem) that has been extensively studied. For this topic, this question is relevant. In particular, a minimal normal subgroup of a supersolvable group must have prime order and by induction on the quotient you can show that "supersolvable" $\implies$ CLT. Finally a group $G$ that is CLT, in particular, has a $p$-complement (i.e. a subgroup with index equal the order of a $p$-Sylow) for every $p$ prime that divides $|G|$, and is solvable by the converse of the Hall's Theorem.
A: Here is a concrete variant of @egreg's answer, using the key fact that $Z(G)$ is not trivial.
If there is an element of order $125$ or $25$, then we're done.
Otherwise, all elements have order $1$ or $5$.
Let $a\in Z(G)$ with $a\ne e$ and $b \in G$ with $b\notin \langle a \rangle$.
Then the set of products $a^i b^j$ is a subgroup of order $25$.
