How to show that a group of order $380$ always has a normal subgroup of $95$

Question

I've already shown in the first part of the question that $G$ has either a normal subgroup of order $5$ or a normal subgroup of order $19$, but I'm not sure how this initial fact means that $G$ (where $G$ is a group of order $380$) always has a normal subgroup of order $95$.

Any help would be much appreciated.

Answer 1

You already have this: If $G$ has more than one $19$-Sylow subgroup, ther must be at least twenty of them, and they are disjoint except for the neutral element. That would account for at least $20\cdot 18=360$ elements of order $19$. Likewise, more than one $5$-Sylow would give us at least $24$ elements of order $5$. As $360+24>380$, we cannot have both, i.e., there must be a unique and hence normal $19$-Slyow or $5$-Sylow.

Now if $H\lhd G$ with $|H|=19$, then the number of $5$-Sylows in $G/H$ is $\equiv 1\pmod 5$ and divides $4$, hence there is one 5-Sylow $N\lhd G/H$. Then $NH\lhd G$ solves the problem.

The argument if $H\lhd G$ with $|H|=19$ is essentially the same.

Answer 2

You have already shown that $G$ has either a normal Sylow $19$-subgroup or a normal Sylow $5$-subgroup. This can be done easily since $n_{19} \in \{1, 20 \}$ and $n_{5} \in \{1, 76 \}$. ($n_{19}=20$ forces $n_{5}=1$ by counting elements.)

Suppose that $G$ has a normal Sylow $19$-subgroup, call it $P$. Look at the group $G/P$. This group has order $20$ and it is equally easy to see that a group of order $20$ always has a normal Sylow $5$-subgroup. If $N/P$ is that normal Sylow subgroup, what is the order of $N$ (which is normal in $G$) ?

Similarly, if $G$ has a normal Sylow $5$-subgroup, say $Q$, consider the group $G/Q$. This group has order $76$ and you can see that a group of order $76$ always has a normal Sylow $19$-subgroup. The preimage of that Sylow subgroup gives you, again, the subgroup you want.