# Understanding the proof that a non-abelian group of order $pq$ ($p<q$) has a nonnormal subgroup of index $q$

I recently tried to come up with my own proof of the following: Let $p$ and $q$ be primes with $p<q$. Prove that a non-abelian group $G$ of order $pq$ has a nonnormal subgroup of index $q$. I found the following proof from this link: Prove that a non-abelian group of order $pq$ ($p<q$) has a nonnormal subgroup of index $q$

Let $K$ be a subgroup of order $q$ of $G$. $K$ exists by Cauchy's Theorem. Then index$[G:K]=p$, the smallest prime dividing $|G|$. Now let $G$ act on the left cosets of $K$ by left-multiplication. The kernel of this action $C=core_G(K)$ is normal in $G$ and $G/C$ injects homomorphically in $S_p$. So $|G/C| \mid p!$. Since $p \lt q$, it follows that $|G/C|=p$, whence $K=C$, and $K$ is normal.

Now $G$ has a subgroup $H$ of order $p$. If it is normal, then we would have $H \cap K =1$ and $|HK|=|H||K|/|H\cap K|=pq$, so $G=HK$, and $G \cong H \times K \cong C_{pq}$, and $G$ would be abelian. So $H$ must be non-normal.

I understand that Cauchy's Theorem implies the existence of $K$, a group of prime order. Since the index $[G:K]=p$ is the smallest prime dividing $|G|$, $K$ is normal in $G$ (Dummit and Foote, page 120). But what's the purpose of the remaining paragraph when it's apparent that $core_G(K)=\bigcap_{g\in G} gKg^{-1}=K$ since $K$ is normal?

And I know that the 1st. isomorphism thm. implies that $|G/core_G(K)|\mid p!$. But I don't think the deduction 'Since $p \lt q$, it follows that $|G/core_G(K)|=p$' is valid: $|G/core_G(K)|=p$ is immediately followed by the fact that $core_G(K)=K$. However, the '$p<q$' seems to be crucial in the proof of the proposition. Just consider $Sym(3)$. It has order $6$, and has $3$ nonnormal subgroups of index $3$, but the only subgroup of index $2$ is normal. So how to appropriately employ the assumption '$p<q$' in the proof?

Another question regarding the second paragraph: 'Now $G$ has a subgroup $H$ of order $p$. If it is normal, then we would have $H \cap K =1$'. Why the 'If it is normal' if $H$ and $K$ have coprime orders implying that $H \cap K =1$?

• what is $Core_{G}(K)$? – Riju Jun 16 '17 at 9:15
• $core_G(K)$ is the kernel of the action of $G$ acting on the set of left cosets of $K$ by left-multiplication, which is the same as the kernel of the corresponding permutation representation afforded by this action. And $core_G(K)=\bigcap_{g\in G} gKg^{-1}$ – user441558 Jun 16 '17 at 9:19

First of all about your last question: $H$ and $K$ being normal has nothing to do with $H \cap K= \{\ e \}\$. The order $H$ and $K$ being co-prime plays the role here. $H$ and $K$ being normal plays the part where you write $HK \cong H \times K$. If both are not normal you won't get a direct product then. Here $K$ which is surely normal of order q and $H$ if normal (of order p) gives you $G \cong C_{p} \times C_{q}$. But if $H$ is not normal, you will get non-abelian group which are actually semi-direct products of $C_{p}$ and $C_{q}$.

Now about the other question about how to effectively use $p<q$ in the proof. See if you use the Sylow theorems then it's quiet evident. But if you don't want to use that there is another easy way to see this.

Use the fact that if there is a unique subgroup of order n in a group, then that subgroup has to be normal.(You should know why?).

Let $K_{1}$ and $K_{2}$ be two distinct subgroups of order q. Then $|K_{1}K_{2}|=q^{2}$(why?). But $q^{2} > pq$ since $p<q$. Hence a contradiction!!

So, You have a unique subgroup of order q and hence normal.

• Why both $H$ and $K$ being normal implies $HK\cong H\times K$, and why if one of them,say $H$, not being normal does not imply $HK\cong H\times K$? In fact, I haven't learnt anything about semi-direct products or Sylow's theorems, but I can understand your approach in the last paragraphs. – user441558 Jun 16 '17 at 9:55
• ok so let's look at the isomorphism.. it's not only H and K being normal also $H\cap K = \{\ e \}\$. Now what is the map you are giving to show $HK \cong H \times K$. – Riju Jun 16 '17 at 10:00
• Oh I see... So we can define $\phi:H\times K\rightarrow HK$ by $\phi(h,k)=hk$ so $ker\phi=\{(h,k)\}$ s.t. $hk=1$. Since $H$ and $K$ have trivial intersection, $h=k=1$. This map is surjective homomorphism so $HK\cong H\times K$. – user441558 Jun 16 '17 at 10:13
• Do you know why is this map well defined? – Riju Jun 16 '17 at 10:17
• if $(h_1,k_1)=(h_2,k_2)$, then $h_1=h_2$ and $k_1=k_2$, so $h_1k_1=h_2k_2$. This map is well-defined. And $\phi((h_1,k_1)(h_2,k_2))=\phi(h_1h_2,k_1k_2)=h_1h_2k_1k_2=...=h_1k_1h_2k_2=\phi(h_1,k_1)\phi(h_2k_2)$. The '$h$' and '$k$' commute because the normalities of $H$ and $K$ imply $[h,k]=hkh^{-1}k^{-1}\in H$ and $\in K$ respectively, so $[h,k]=1$ since $H$ and $K$ have trivial intersections. So $hk=kh$. This took me a whole 15 minutes to figure out! – user441558 Jun 16 '17 at 10:41

The purpose of the rest of the first paragraph is to prove the special case (perhaps) needed here of the statement from Dummit and Foote that you mention.

In fact, this argument proves it in general. We have $$p[K:C] = [G:K][K:C] = [G:C] | p!, \quad \text{hence} \quad [K:C]|(p-1)!.$$ Since $|K|$ divides $|G|$, it cannot be divisible by any prime number smaller than $p$. This proves that $[K:C] = 1$, hence that $K = C$.

The "if it is normal" assumption isn't really needed to prove that $H \cap K = \{1\}$. It's only used to go from $G = HK$ to $G \cong H \times K$.

• This is a bit awkward... I really thought that the first paragraph was some circular reasoning crap :P Never knew that it was to prove a special case of THAT theorem. – user441558 Jun 16 '17 at 10:44