Show that every group of order 100 has a subgroup of order 50. Show also that the number of subgroups of order 50 is either 1 or 3.

For the first part I did the following:

As $|G|=100=2^2 5^2$ we can deduce from Sylow's Theorems that $G$ has a normal subgroup $N$ of order 25 (normal because there is only one). There is also a subgroup $H$ of order 2 (as 2 divides 100). Since $N$ is normal there holds $NH=HN$. So $NH$ is in fact a subgroup, which has order 50.

But I am puzzled as to why there are either 1 or 3 such subgroups.


By Correspondence Theorem, the subgroups of $G$ containing $N$ are in one-to-one correspondence with the subgroups of $G/N$, with the map given by $H\mapsto H/N$. Since $|G/N|=4$, then there exist either $1$ or $3$ subgroups of $G/N$ of order $2$ (why?), which means that there are either $1$ or $3$ subgroups of $G$ containing $N$, aside from $N$ and $G$.

It remains for you to show that if $K$ is a subgroup of $G$ of order $50$, then $K$ contains $N$. (This isn't tricky.)

  • 1
    $\begingroup$ There exist either 1 or 3 subgroups of $G/N$ of order 2, because $G/N$ is either $V_4$, in which case it has 3, or it is $C_4$ and then it has only 1 subgroup of order 2. And if $K$ is a subgroup of $G$ of order 50, then it contains a subgroup of order $5^2=25$, but $N$ is the only subgroup of order 25 (as mentioned above). $\endgroup$ – Phil-ZXX Apr 23 '13 at 0:43
  • $\begingroup$ Precisely so, Thomas. $\endgroup$ – Cameron Buie Apr 23 '13 at 0:50
  • 2
    $\begingroup$ @Thomas Another 'first principles' argument that doesn't require enumerating the groups of order 4: if $g\in G/N$ has order 4 then $g^2$ has order 2, so at least one element in $G/N$ has order 2. It can't be more than 3 (there are only four elements and one of them has order 1!), and if $g$ and $h$ are distinct elements of order 2, then $gh$ is also of order 2 (why?) and can't be either $g$ or $h$ (again: why?) so it's a third element of order 2. $\endgroup$ – Steven Stadnicki Apr 23 '13 at 1:10
  • $\begingroup$ $gh$ is also of order 2, because $(gh)^2=ghgh=g^2h^2=ee=e$ (since groups of order 4 are abelian) and if $gh$ was, say, $g$, then $g=gh\Rightarrow h=e$, but $e$ has order 1. $\endgroup$ – Phil-ZXX Apr 23 '13 at 1:17
  • $\begingroup$ Nice one, @Steven! $\endgroup$ – Cameron Buie Apr 23 '13 at 1:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.