A normal subgroup so that any homomorphism into a $p$-group is trivial on it. 
Problem
  Let $G$ be a finite group of order $n$ and $p|n$. Show that there is a unique normal subgroup $N$ satisfying the following property:



*

*$G/N$ is a $p$-group (I guess it can be trivial group);

*any homomorphism $\pi$ of $G$ into a $p$-group is trivial on $N$ i.e. $\pi (N)=1$.

My idea is to consider the set of all elements whose order cannot divided by $p$, then any element in this set will be sended to the identity by the above $\pi$
 A: Hmm.  Not sure this is true.  Consider $A_5$ which is a simple group of order $60$.  Since the only normal subgroups of $A_5$ are $\{e\}$ and $A_5$, and neither of the quotient groups are $p$ groups.
A: Suppose $G/N$ is a $p$-group; then if $x\notin N$ and $x^m\in N$ with $m$ minimal we must have that $m$ is a power of $p$, because if $\phi:G\to G/N$ is the surjective homomorphism with kernel $N$ then $\phi(x)\neq e$ and $\phi(x^m)=\phi(x)^m=e$, and $m$ is minimal, meaning $m$ is the order of $\phi(x)$. This criterion is equivalent to $G/N$ being a $p$-group, as it means every element of $G/N$ is a power of $p$.
Suppose $N_1,N_2$ are normal subgroups such that $G/N_1$ and $G/N_2$ are $p$-groups. I claim that $G/(N_1\cap N_2)$ is a $p$-group. To see this, suppose $y\notin N_1\cap N_2$ and let $m$ be the smallest positive integer such that $y^m\in N_1\cap N_2$. If $y\in N_1-N_2$, then $y\notin N_2$ and $m$ is the smallest positive integer such that $y^m\in N_2$, and since $G/N_2$ is a $p$-group this means $m$ is a power of $p$. A similar argument shows $m$ is a power of $p$ if $y\in N_2-N_1$. Suppose then that $y\notin N_1\cup N_2$. Let $r$ be the minimal positive integer such that $y^r\in N_1$ and let $s$ be the minimal positive integer such that $(y^r)^s\in N_2$. Then if $\psi:G\to G/(N_1\cap N_2)$ is the surjective homomorphism with kernel $N_1\cap N_2$ we must have that $\psi(y)^{rs}=e$, so the order of $\psi(y)$ divides $rs$, which is a power of $p$. Since $m$ is minimal and $\psi(y)^m=e$, this implies that $m$ is a power of $p$.
Now we can show that there is a unique minimal normal subgroup $N$ such that $G/N$ is a $p$-group by letting $N$ be the intersection of all normal subgroups $N'$ such that $G/N'$ is a $p$-group. By definition, for all $N'$ such that $G/N'$ is a $p$-group we must have that $N\leq N'$. Thus if we have any homomorphism $\phi:G\to P$ with $P$ a $p$-group we must have that $\phi(N)=\{e\}$, so $N$ satisfies the desired conditions. To see that $N$ is unique, let $f:G\to G/N$ be a surjective homomorphism. If $N'$ is such that $G/N'$ is a $p$-group and $N'\neq N$, then there exists an element $z\in N'-N$ and hence $f(z)\neq e$, and we are done.
