Assume a finite group $G$ with $|G| = 40$. Show that the subgroup of
order $8$ is normal and unique.
Attempt:
Since $|G| = 2^3 \cdot 5$, by the $1$st Sylow theorem, $G$ has at least one Sylow $2$-subgroup of order $8$.
Now, using the $3$rd Sylow theorem, the number $N$ of those Sylow $2$-subgroups is an odd number and divides $40$.
Since $1,2,4,5,8,10,20$ are the only divisors of $40$, smaller than $40$, those Sylow $2$-subgroups can either be $1$ or $5$.
If $N = 1$, it can be easily shown that this unique Sylow $2$-subgroup is normal and we're done.
My problem lies in the $N = 5$ case:
Assume that there exists $5$ Sylow $2$-subgroups of order $8$ and let $H,K$ be two of them.
Then, because
$$
|HK| = \frac{|H||K|}{|H \cap K|}
$$
$|H \cap K|$ must have at least $2$ elements. If it didn't, $|HK|$ would have $64$ elements, which is a contradiction.
Therefore $N[H \cap K]$'s order is a multiple of $8$ and a divisor of $40$. That leaves us with $|N[H \cap K]| = 40$ an thus:
$$
H \cap K \trianglelefteq G
$$
Is there a mistake somewhere? I cannot see why $5$ Sylow $2$-subgroups cannot coexist within $G$.
Best Answer
To support the argument in the comment by @the_fox , I am posting a proof.
Consider the dihedral group $D_{40}$ with the generators $r,s$ satisfying $o(r)=20,o(s)=2$ and $srs=r^-$
$D_{40}=\{1,s,sr^i,r^i : 1\le i \le 19\}$
Then consider the elements $x=s,y=r^5$, then $xyx=sr^5s=(srs)^5=(r^-)^5=(r^5)^-=y^-$.
Also $o(x)=2,o(y)=4$ and thus $H=\langle x,y\rangle$ is 8 order subgroup of $G$
Now let $g=sr$ and $y$ be the same as above. Then $o(g)=2$ and
$gyg=srr^5sr=sr^6sr=(srs)^6r=r^{-6}r=(r^5)^-=y^-$
Thus $K=\langle g,y \rangle$ is again a subgroup of order $8$
Now these are not the same subgroups as $sr^6 \in K$ but $ sr^6 \notin H$ which can be easily checked . Since there are more than one $2$-Sylow subgroup, it is not normal.
The various subgroups of eight elements are the groups of symmetries of the five different squares (shown in different colors) inscribed in a regular 20-gon. The big group permutes the five squares, and the Sylow $2$-subgroups are the stabilizers.