9
$\begingroup$

If $o(G)$ is $pq$, $p>q$ are primes, prove that $G$ has a subgroup of order $p$ and a subgroup of order $q$.

[This question is from Herstein and it comes before Sylow's and Cauchy's theorem. So I'm expecting an answer without using any of these]

Here's what I got so far:

If $G$ is cyclic then we are done otherwise, we can assume that it is not cyclic which means every non-identity element must be of order $p$ or $q$.

Case $(1)$ if there exists $a\in G$ such that $o(a) = p$ and if there also exists an element of order $q$ then we are done. So we can assume that every non-identity element is of order $p$. Now pick $b\in G$ such that $b\notin \langle a \rangle$ then $o(b) = p$ and $\langle a \rangle\cap\langle b \rangle =(e)$

So we have $\langle a\rangle \langle b\rangle\subset G$ but $o(\langle a \rangle \langle b \rangle) = \dfrac {o(\langle a \rangle)o(\langle b \rangle)}{o(\langle a\rangle \cap \langle b\rangle)} = p^2$ but $p^2 > pq$ [since $p>q$] so we got a contradiction.

Give me a hint for the second case and correct me if my argument for the first case is wrong

$\endgroup$
16
  • 1
    $\begingroup$ In general $\langle a\rangle\langle b\rangle$ is not a subgroup, but the subgroup generated by them has order at least what you suggest. $\endgroup$ Aug 16 '20 at 17:15
  • 2
    $\begingroup$ I didn't say that it's a subgroup, I only used it as a subset and that formula works for the product of two finite subgroups @DavidA.Craven $\endgroup$ Aug 16 '20 at 17:19
  • $\begingroup$ Your argument in the first case looks good to me. $\endgroup$ Aug 16 '20 at 17:22
  • $\begingroup$ @LucasHaobam OK, yes. It's just a subset. It's fine, but is not the way you are supposed to do it, as it doesn't work for the other prime. $\endgroup$ Aug 16 '20 at 17:38
  • $\begingroup$ Okay, give me a hint. I'll try @DavidA.Craven $\endgroup$ Aug 16 '20 at 17:41
5
$\begingroup$

Assume that every non-identity element generates a cyclic group of order $q$, the smaller of the primes.

Conjugacy is an equivalence relation on a group. So, we should be able to partition the group into its equivalence classes. The size of the equivalence class an element belongs to is the index of the centralizer of the element. Why? Fix $x\in G$. Make a homomorphism from $G \rightarrow G$ by sending $g \rightarrow xgx^{-1}$. The size of the equivalence class is the order of the image. What is the kernel of this map?

If the centralizer is of order $p$ or $pq$, we are done. Assume every centralizer is of order $q$, the index of the centralizer is $pq/q=p$. Every element would belong in a equivalence class of size $p$, except for the identity element.

A simple cardinality calculation shows that $pq= kp+1$, where represents the number of equivalence classes. However, this is absurd and therefore, not every subgroup of order $q$.

$\endgroup$
1
  • 1
    $\begingroup$ This works for both cases $\endgroup$ Aug 16 '20 at 21:20
0
$\begingroup$

Firstly, if all the non-trivial elements of $G$ have order $pq$, then, for any given $g\in G\setminus\{e\}$, we get $e=g^{pq}=(g^p)^q$ and hence $o(g^p)\mid q$; but $g^p\in G\setminus\{e\}$, so, by assumption, $o(g^p)=pq$ and we end up with $pq\mid q$: contradiction. Therefore, there must be elements of order $p$ and/or elements of order $q$. If $G$ is Abelian, then your argument works for the "second case" either, because then every subgroup is normal and hence $\langle a\rangle\langle b\rangle$ is a subgroup of $G$ such that $o(\langle a \rangle \langle b \rangle) = \dfrac {o(\langle a \rangle)o(\langle b \rangle)}{o(\langle a\rangle \cap \langle b\rangle)} = q^2\nmid pq$: contradiction (Lagrange). So, there must be elements of both orders, $p$ and $q$. If $G$ is nonabelian, then it has trivial center (see here) and the Class Equation reads: $$pq=1+k_pp+k_qq$$ where $k_i$ are the number of conjugacy classes of size $i$. None of the $k_i$ can be zero, because neither $p\nmid 1$ nor $q\nmid 1$. Say $O_p$ a conjugacy class of size $p$ and $O_q$ a conjugacy class of size $q$, which as said both exist; for $g\in O_p$, we get $|C_G(g)|=q$ (Orbit-Stabilizer Theorem) and hence $\langle g\rangle=C_G(g)$ (because in general $\langle g\rangle\le C_G(g)$) and finally $o(g)=q$. Likewise, for $h\in O_q$, we get $|C_G(h)|=p$ and hence $\langle h\rangle=C_G(h)$ and $o(h)=p$.

$\endgroup$
5
  • $\begingroup$ You are trying to prove that $G$ has elements of orders $p$ and $q$. Certainly if $G$ has an element of order $pq$ then powers of it have order $p$ and $q$. So your only issue is if all non-trivial elements have order $p$, or all have order $q$. Which you don't appear to address. $\endgroup$ Jul 30 at 10:22
  • $\begingroup$ @David A. Craven, The OP has proved that not all the non-trivial elements have order $p$. With a similar argument, in the case of Abelian $G$, I (think to) have proved that not all the non-trivial elements have order $q$. So the Abelian case is done. Then, I dealt with the nonabelian case separately, as a whole, with the Class Equation. Is my logic flawn somewhere? $\endgroup$
    – CAB
    Jul 30 at 10:44
  • 1
    $\begingroup$ The main part of your solution is the statement that, for a non-abelian group, the centralizer of an element of order $p$ has order $p$, and similarly for $q$. Once you state that, which is just half a line, then everything else follows immediately. You seem to have buried the pertient part with lots of other things. For the abelian case, the proof is very fast. $\endgroup$ Jul 30 at 11:44
  • $\begingroup$ Here is the proof. Assume $p\neq q$, but to make statements quicker, do not assume $p>q$. If $G$ has an element $x$ of order $pq$ then $x^p$ has order $q$, $x^q$ has order $p$, and done. So all elements have order $1$ or $p$, or $1$ and $q$. Let $x$ be an element of order $p$. If $G$ is abelian then $H=\langle x\rangle$ is normal, and $G/H$ has order $q$. Let $Hy$ have order $q$ in $G/H$. Then $y$ cannot have order $p$, because its image has order $q$. Done. $\endgroup$ Jul 30 at 11:47
  • $\begingroup$ If $G$ is non-abelian then $G/Z(G)$ cannot have prime order, and $G\neq Z(G)$, so $Z(G)=1$. In particular, any element of order $p$ has centralizer of order exactly $p$. Thus if $G$ has no elements of order $q$ and $c$ classes of order $p$ (each of which as size $q$), then $|G|=pq=1+cq$, which is impossible. $\endgroup$ Jul 30 at 11:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.