prove homomorphic image has order 4 Let $G$ be a group of order $20$ show that $G$ has homomorphic image of order $4$. 
From Cauchy theorem we have elements $a,b \in G$ of orders $2,5$ respectively then 
from first isomorphic theorem homomorphic image is isomorphic with $G/kerf$ for some homomorphism $f:G \to im(f)$
so if $kerf=<b>$ then will be ok but to show that I need $<b>$ is normal subgroup of $G$ and I have some problems with that

$G/\langle b\rangle$ is such a homomorphic image. The subgroup generated by $b$ is normal by the Sylow's theorems. Indeed the number $n$ of subgroups of order $5$ is congruent to $1\mod 5$, and a divisor of $20/5=4$. Hence $n=1$, which proves the subgroup is normal.
