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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 $G$.

Shaun
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Andrew
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    That's not true. What about the dihedral group of order 40? What is true (and perhaps this is what the question was really about) is that there is only one Sylow $5$-subgroup. – the_fox Jun 15 '20 at 19:48
  • The equation $|HK|=|H||K|/|H\cap K|$ is not in general true. – Andreas Blass Jun 15 '20 at 23:31
  • @AndreasBlass Is it not? We have $h_1k_1=h_2k_2$ if and only if $h_2^{-1}h_1=k_2k_1^{-1}\in H\cap K$, and the formula follows from that. For me $$HK={hk\in G\mid h\in H, k\in K}.$$ The trouble being that in general $HK$ need not be a subgroup. If you define $HK$ as the subgroup generated by $H\cup K$ then I agree. – Jyrki Lahtonen Jun 16 '20 at 05:27
  • @JyrkiLahtonen Yes, I understood $HK$ to mean the subgroup generated by $H$ and $K$. – Andreas Blass Jun 16 '20 at 17:13

1 Answers1

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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.

enter image description here

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.

verret
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