# Assume $|G|=40$. Show that the subgroup of order $8$ is normal and unique.

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 $$G$$.

• 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. Jun 15 '20 at 19:48
• The equation $|HK|=|H||K|/|H\cap K|$ is not in general true. 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. Jun 16 '20 at 5:27
• @JyrkiLahtonen Yes, I understood $HK$ to mean the subgroup generated by $H$ and $K$. Jun 16 '20 at 17:13

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.

• +1 I took the liberty of adding a picture that hopefully makes it clearer to a casual reader what's going on. If you object, feel free to remove it. Jun 15 '20 at 22:06
• @JyrkiLahtonen..That is enlightening !! Thanks for the picture. Jun 16 '20 at 7:09