A finite non cyclic group, all of whose proper subgroups are cyclic, has a non trivial proper normal subgroup. 
Theorem: A finite non cyclic group, all of whose proper subgroups are cyclic, has a non trivial proper normal subgroup.

Proof : Suppose there are no non trivial proper normal subgroup. 
Claim 1 :If $H$ and $K$ are two distinct maximal subgroups then 
$H\cap K=\{e\} $. 
Proof of claim 1: 
We have 
$H,K \subsetneq N(H\cap K) $
since $H$ and $K$ are cyclic by hypothesis.Now $ N(H\cap K)=G$ since $H$ and $K$ are maximal subgroups. So $H\cap K \triangleleft G $, but by assumption there is no non-trivial normal subgroup .So $H\cap K=\{e\}$
Thus our claim is proved.
Let $H$ be a fixed maximal subgroup of G . Then there exist $x\notin
 H$ such that $xHx^{-1}\neq H$ otherwise if  no such $x$ exist then $H\triangleleft G$ , a contradiction to our assumption that no proper normal subgroup exist.
Claim 2: $xHx^{-1}$ is maximal. Otherwise 
$xHx^{-1}\subsetneq K\subsetneq G$ for some K but then
$H \subsetneq x^{-1}Kx\subsetneq G$ , a contradiction that H is maximal.
So by claim 1,$H\cap xHx^{-1}=\{e\}$.
Since $H$ is not normal in G and is maximal, so $ N(H)=H$.
The no. of non-identity elements of H and it's conjugate is 
$(|H|-1)[G:N(H)] \quad $.... ($A$)
Now $(|H|-1)[G:N(H)]=(|H|-1)[G:H]=|G|-|G|/|H|\ge |G|/2 $
Again since $[G:H]> 2 $ ($H$ being not normal) , the no. of elements of H including identity and their conjugates is $ |G|-[G:H]+1< |G|-1$
So there exist $y$ not in $H$ and any of it's conjugates.Let $K$ be a naximal subgroup containing $y$.
Proceeding as earlier ,there exist at least $|G|/2 $ non identity elements and their conjugates in $K$ .
Hence in total, we have $|G|$ many non identity elements in $G$, which is contradiction.... $\quad ($B$)$. 
So our supposition is false, there is a proper normal subgroup.
I am having problem in statement $(A)$ and $(B)$. What is the proof of statement $(A)$? Is it true for any $H\le G$ ( I mean true, in general). In statement $(B)$ , why is there no  non trivial intersection between the conjugates of elements of $H$ and $K$ ? . That will reduce the counting of non identity elements in G. Please help me in understanding the concept. 
 A: Every conjugate of $H$ contains $|H|-1$ non-identity elements and every non-identity element conjugate to one in $H$ lies in exactly one conjugate of $H$ by Claim 1. The number of conjugates of $H$ is $[G:N(H)]$ and so there are $(|H|-1) [G:N(H)]$ elements conjugate to an element of $H$.
In other words (or equations), Claim 1 implies that you have 
$$
|(\bigcup_{g \in G} {}^g H) - \{1\} | \\
= |\biguplus_{g N(H) \in G/N(H)} ({}^g H - \{1\}) | \\
= \sum_{g N(H) \in G/N(H)} |{}^g H - \{1\}| \\
= \sum_{g N(H) \in G/N(H)} (|H| - 1) \\
= (|H|-1) [G:N(H)] 
$$
Now, for (B), if you have $y \in G$ which does not lie in any conjugate and $K \subseteq G$ a maximal subgroup of $G$ which contains $y$, then $K$ is not conjugate to $H$ (otherwise $y$ would lie in a conjugate of $H$) and thus any conjugate of $K$ intersects any conjugate of $H$ trivially by Claim 1 (as with $H$, all the conjugates of $K$ are maximal subgroups of $G$).
It follows that $(\bigcup_{g \in G} {}^g H) - \{1\}$ and $(\bigcup_{g \in G} {}^g K) - \{1\}$ are disjoint and by the calculation before, both of these subsets of $G$ contain at least $|G|/2$ elements, this adds up to at least $|G|$ elements, but the identity is also missing and this gives the desired contradiction.
