How many subgroups of order $3$ does a non-abelian group of order $39$ have? How many subgroups of order $3$ does a non-abelian group of order $39$ have ? 
My work : let $n_{13}$ and $n_3$ denote the number of sylow subgroups of a non-abelian group $39$ have .
Then $n_{13}$ is not zero by Sylow's theorem and $n_{13}|3 , n_{13}=1(\mod 13) $ . This two equation give solution $n_13= 1 $ . So $G$ has exactly  one subgroup of order $13$ . 
Now $n_3|13 , n_3=1(\mod 3) $ . This two equations give solutions $n_3=1 or 13 $. 
Assume $n_3=13$ . Then there are $1+(13-1)3=27$ elements  order $3$ . So Now we have $27$ elements of order $3$ and $1$ element of order $13$ . But the group has $38$ non-identity element . Sine the only possible order of the non-identity elements of $G$ are $3$ and $13$ . This shows lack of lack enough group elements . Hence $n_3\neq 13$ . So $n_3$ must be $1$ .
Is my solution correct ?If not you are welcomed to provide a solution . Thank you .
 A: If $n_3=13$, then there should be $13*(3-1)=26$ elements of order $3$. This is because there are $13$ distinct subgroups of order $3$ while each subgroup has $2$ elements of order $3$. Also, note that each pair of subgroups of order $3$ intersect trivially.
If $n_3=1$, then $G$ has a normal subgroup of order $3$,say $P$.
Note that $G$ has also a normal subgroup of order $13$,say $Q$.
Hence $G=P\times Q$ which is abelian since $P$ and $Q$ are both abelian, a contradiction. 
So $n_3$ must be $13$ and the number of elements of order $3$ are $26$. 
A: Alan's proof is perfectly fine. But if you want to go by an order argument. Then as you have shown $n_{13}$=1. So there are 12 element of order 13. If $n_3$ would have been 1 you would get 2 elements of order 3. But then there are more elements remaining, whence there should be an element of order 39 which will make G cyclic and hence commutative. So only choice is $n_{3}=13$ and hence you have 26 elements of order 3.
For a general information,  A group of order 39 is either cyclic or else there is a unique non-abelian subgroup of order 39 upto isomorphism.
A: We know that there is only one subgroup of order $13$(By Sylow's thm) which implies there are exactly $12$ elements of order $13$ (precisely the non-identity elements of the subgroup of order $13$).
Now every element has either order=$3$ or order=$13$ or order=$1$ (by Lagrange's thm).
No of elements of order $1$ is $1$ which is the identity of the group which is unique.
So all the other elements are of order $3$ so the ans is $39-12-1=26$
