3

Prove that a group of order $75$ can have at most one subgroup of order $25$.

My attempt (Proof by contradiction) Suppose $H$ and $K$ are two subgroups of order $25$. $ | H \cap K | $ divides $25$, $ | H \cap K | $ can be $1$ or $5$. Therefore $|HK| = |H||K|/ | H \cap K | = 625$ or $125$.

I am lost after this. Any help or suggestion will be much appreciated.

I am trying to solve without using Sylow's theorem.  

citadel
  • 2,940
Bluey
  • 2,114
  • You can prove that HK is a group, because H is normal subgroup of G, so HK=KH. Through this you can reach contradiction. – Shak Jun 12 '18 at 07:11

2 Answers2

7

Hint: $HK$ is a subset (although not necessarily a subgroup) of the original group. That limits how large it can be.

Arthur
  • 199,419
1

With somewhat more sophistication, if $|G|=75$ and $H \lt G$, with $|H|=25$, then $|G:H|=3$, the smallest prime dividing the order of $G$. Hence $H \lhd G$, see here for my simple general proof (which you will not find easily in textbooks) along the same lines as your counting/reasoning above.

Now let $K \lt G$, also with $|K|=25$. Then look at the image of $K$ in the quotient group $G/H$ and use the second isomorphism theorem: $KH/H \cong K/(K \cap H)$. $|KH/H|$ divides $|G/H|$ and $|K/(K \cap H)|$ divides $|K|=|H|$. But gcd$(|H|,|G:H|)=1$, so $KH/H=\bar{1}$ (and $K/(K \cap H)=\bar{1}$), whence $K \subseteq H$ and since both subgroups have equal order, $K=H$. Hence, $H$ is unique.

So what we see here in general, if $N \lhd G$, gcd$(|N|,|G:N|)=1$, then $N$ is a unique subgroup of order $|N|$. And this implies that it is characteristic (fixed by every automorphism of $G$). As an example: a normal Sylow subgroup satisfies these properties.

Nicky Hekster
  • 49,281