Does the question that every group of order 77 must have an element of order 7 and an element of order 11 have an elementary proof

Question

Question 21 of Chapter 10 of Abstract Algebra by Dan Saracino is as follows:

Prove that every group of order 77 must have an element of order 7 and an element of order 11.

The more general case of this question is to prove a group of order $pq$ must have an element of order $p$ and an element of order $q$, where $p$ and $q$ are distinct primes.

I have seen this question on this site with so many solutions. However, none of them works for me because, up to Chapter 10, Sylow Theorem, Cauchy Theorem, homomorphism, quotient group, and even normal subgroups have not been introduced yet. Hence, there should be an elementary proof for Question 10.21 of Dan Saracino's book, but I do not know whether this elementary proof can also work for the general case or not.

I have a proof for the general case, but I know there is a problem with this proof that I will mention after the proof:

My proof

Suppose $G$ is our group such that $|G|=pq \ ;\ p>q$. There are two possibilities:

1. $G$ is cyclic.
2. $G$ is not cyclic.

In the first case, we can consider $G=\langle x\rangle $, so $\langle x^p\rangle $ and $\langle x^q\rangle $ are subgroups of order $q$ and $p$, respectively.

In the second case, we have 3 possibilities:

1. All elements have order $p$.

2. All elements have order $q$.

3. Some elements have order $p$ some others have order $q$.

Now, we are showing that the first and second possibilities in this case can not exist, and so only the third one is acceptable.

If all elements are of order $p$:

There is an element $x$ with order $p$. $G= \{e,x,x^2,\cdots,x^{p-1}, \cdots \}$. Now, we only know p elements of this group. What about the other elements? So, we name another element $y$ such that $y \not\in \langle x\rangle $. According to the assumption, the order of $y$ should be $p$. So $G=\{e,x,x^2,\cdots,x^{p-1},y,y^2,\cdots,y^{p-1} \cdots \}$. You should know that $\langle x\rangle \cap \langle y\rangle = \emptyset$. We know that the group should contain the product of every two elements, and since $x^iy^j \ne x^ry^s$ for all $i \ne r $ or $ j \ne s$, $G$ should contain at least $p^2$ elements. This is a contradiction because $p^2 > pq$.

If all elements are of order $q$:

With a same argument, we can say $G=\{e,x,x^2,\cdots,x^{q-1},y,y^2,\cdots,y^{q-1} \cdots \}$. Now, $G$ should contain $q^2$ elements which are the product of elements of $\langle x\rangle $ and $\langle y\rangle $. These $q^2$ elements form a subgroup in $G$. Indeed the subgroup is denoted by $\langle x,y\rangle $ (i.e. elements which are generated by $x$ and $y$). $|\langle x,y\rangle|=q^2$, but, according to the lagrange theorem, since $q^2$ does not devide $pq$, this is impossible to have such a subgroup, and we reach a contradiction.

Hence, $G$ should contain both subgroups of order $p$ and $q$.

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The problem with this proof is that, unless I know elements of $\langle x\rangle $ can commute with elements of $\langle y\rangle $, I cannot say $|\langle x,y\rangle|=q^2$.

Now, can anyone say how I should deal with this problem?

If not, can anyone give me an elementary proof for the general case without using Sylow Theorem, Cauchy Theorem, homomorphism, quotient group, and normal subgroups?

If not, can anyone give me an elementary proof for the case when $|G|=77$ without using Sylow Theorem, Cauchy Theorem, homomorphism, quotient group, and normal subgroups?

Answer 1

I suppose the following argument can be an elementary proof:

First, according to Question 10.27 of Dan Saracino's book, I use the following lemma.

Lemma:

If $G$ is a finite group, $[G: Z(G)]$ cannot be prime.

Proof:

If $Z(G)=|G|$, then $[G: Z(G)]=1$, which is not prime. Otherwise, there exists $x \in G $ and $ \not \in Z(G)$. Hence, $Z(G) \subset C_G(x) \subset G$. Since $x \not \in Z(G)$, we have $Z(G) \ne C_G(x) \ne G$. Therefore, by Lagrange Theorem, there exist $a,b \ne 1$ such that $ a|Z(G)| = |C_G(x)|$ and $ b|C_G(x)|=|G|$. So, $[G: Z(G)]=ab$ which is not prime.

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According to the lemma, when $|G|=pq$, $|Z(G)|=1$ or $G$ is abelian.

1- If $|Z(G)|=1$:

By class equation, $pq=1+pq(\frac{1}{|C_G(a_1)|}+\frac{1}{|C_G(a_2)|}+ \cdots +\frac{1}{|C_G(a_m)|})$, where $|C_G(a_i)|=p$ for some $i$ and $|C_G(a_i)|=q$ for some others.

If there is no ellements of order $p$ or $q$, then $|C_G(a_i)|=q$ for all $i$ or $|C_G(a_i)|=p$ for all $i$, respectively. Hence, $pq=1+\alpha p$ or $pq=1+\alpha q$. Since $p(q-\alpha)=1$ or $q(p-\alpha)=1$ are not possible, $G$ should contain elements of both order $p$ and $q$.

2- If $G$ is abelian:

Soppose $|G|=pq\ ;\ p>q$. If all elements have order $p$ then there exist $x,y \in G$ such that $\langle x\rangle \cap \langle y\rangle =\{e \}$. Therefore |G| should be at least $p^2$. Since $p^2 > qp$, all elements can not have order $p$.

If all elements have order $q$ then there exist $x,y \in G$ such that $\langle x\rangle \cap \langle y\rangle =\{e \}$. Therefore $H=\{e,x,\cdots,x^{q-1}, y,\cdots,x^{q-1}y,y^2,\cdots,x^{q-1}y^2,\cdots,y^{q-1},\cdots,x^{q-1}y^{q-1} \}$ is a subgroup of order $q^2$, but, since $q^2$ does not devide $pq$, this is impossible. Hence, all elements can not have order $q$, and $G$ should contain elements of both order $p$ and $q$.

I think this is the correct answer. If it is not, I appreciate it if you tell me in the comments.

Answer 2

There are two things wrong with your proof as it stands - first two subgroups of the same group can never have an empty intersection - the identity will be an element of both.

Second, there are groups of order $pq$ which are not commutative (the smallest has order $21$, and it happens whenever $p\gt q$ and $q$ is a factor of $p-1$ (so here $3$ is a factor of $7-1=6$), so if you want elements to commute, you need to prove it.

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The way to do this is via a rather better counting argument.

The consequence of every element other than the identity having order other $p$ is that each element of order $p$ is in a cyclic subgroup of order $p$ and this subgroup has $p-1$ non-identity elements. These subgroups are all distinct and if there are $n$ of them we can count elements to get $$pq=n(p-1)+1$$ so that $p(n-q)=n-1$

Now we can choose the higher prime for $p$ so that $p\gt q$.

It is easy to see successively $n\gt 1$, $n\gt p$ (because $p$ is a factor of $n-1$) but this means also that $pq\gt p^2-p+1$ so that $q\gt p-1$ and this leads to a contradiction.

Sylow's theorems are essentially counting theorems, using the properties of prime numbers - and there are several other significant counting arguments in group theory.

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[Added in response to comments]

Now suppose $p\lt q$. Then we have more cases.

We can have $p=2$ so the order of the group is $2q$ and it is easy to show that if all elements have order $2$ then the group is commutative and if finite has order a power of $2$.

If $p\gt 2$ then $n\gt 1$ and therefore $n\gt q$ and $n=mp+1$ for some integer $m\ge 2$.

We then have $q-1=m(p-1)$

And I think we probably have to do something approximating a proof of Sylow to show that this doesn't work, as here the arithmetic runs out.

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Note that if $p=q$ we scrape by with $n=p+1$ and $p^2=(p+1)(p-1)+1$

Answer 3

The simple counting argument from Mark Bennet's answer works fine in the specific case of $\vert G\vert=77$.

Suppose a group $G$ has only elements of order $a$. Each cyclic subgroup has $a-1$ non-identity elements, so $\vert G\vert=n(a-1)+1$, where $n$ is the number of distinct cyclic subgroups. That is, $\vert G\vert \equiv 1$ modulo $a-1$.

In the case $\vert G\vert=77$, we have $77\not\equiv 1$ modulo 10 or modulo 6. Thus we cannot have only elements of order 11 or only elements of order 7.