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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.

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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.)

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    $\begingroup$ 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
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    $\begingroup$ @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

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