Proving that a group of order $77$ is cyclic.

Question

Prove that a group of order $77$ is cyclic.

I reached a step then got stuck.

My attempt:

Let $G$ be a group with $|G|= 77$. $G$ may have elements of orders $7$, $11$ and $77$ (divisors of $77$).

If $G$ has an element of order $77$, then we are done.

If $G$ has only elements of order $7$, then the number of elements of order $7$ is divisible by $\phi(7)= 6$, then we will have, due to the presence of the identity element of order $1$, $|G|= 77= 6k+1$, for some $k$. This yields that $76=6k$, but $6\nmid 76$.

Similarly if we suppose that $G$ has only elements of order $11$. We will work the same way until reaching $10\nmid 76$.

Hence we conclude that $G$ has elements of order $7$ and $11$.

Here I don't know how to continue.

I know that if $a,b \in G$, where $|a|= 7$ and $|b|= 11$ then $|ab|$ divides $lcm(7,11)= 77$, but how to show that there is an element of order $77$??

I should let you know that I didn't take the theorem saying:

$|HK|= \frac{|H||K|}{|H\cap K|}$, as I see it in similar proofs.

Answer 1

We're given that $|G| = 77 = 7 \cdot 11$. Since $7, 11$ are prime and $7 \not\mid (11 - 1)$, Sylow's Third Theorem implies that $G$ has exactly one subgroup of order $7$ and one of order $11$. So, $G$ contains exactly $7 - 1 = 6$ elements of order $7$, $11 - 1 = 10$ elements of order $11$, and $1$ element of order $1$ (the identity). Since $1 + 6 + 10 = 17 < 77$, $G$ must have elements with some order $\neq 1, 7, 11$, hence an element (in fact, $60$ elements) of order $77$.

The same argument applies mutatis mutandis for any group whose order is a product $p q$ of primes $p < q$ for which $p \not\mid (q - 1)$.

Answer 2

$|G|$=$77$ By Sylow's First Theorem,

Let $n_{11}$ be the number of Sylow-11-Subgroup of order $11$

Then $n_{11} =11k+1$ for some k $\in$ Z

And $n_{11}$$\mid$$77$

If you check carefully then $k=0$ is the only solution. Hence, the Sylow-11-Subgroup is normal. Let $H$ be the unique Sylow-11-Subgroup.

Similarly, check for Sylow-7-Subgroup. You will find it is also normal. Let $K$ be the unique Sylow-7-Subgroup.

Now, $H$$\cong$$Z_{11}$ so it is a cyclic group of order 11 and $K$$\cong$$Z_{7}$ so it is cyclic group of order 7.

$H$ and $K$ are normal and $gcd(7,11)=1$

So, $G$=$H$×$K$=$Z_{11}$×$Z_{7}$=$Z_{77}$

So,$G$$\cong$$Z_{77}$ and so it is a cyclic group of order 77

Answer 3

An argument without Sylow and Cauchy goes as follows.

For $G$ a group of order $pq$, with $p,q$ primes (WLOG $p<q$), the center $Z(G)$ is trivial or equal to the whole $G$. In fact, suppose $|Z(G)|=p$; then there is some $x\in G\setminus Z(G)$, and hence $Z(G)<C_G(x)<G$; but then (Lagrange) $p\mid |C_G(x)|$ and $|C_G(x)|\mid pq$; namely, $|C_G(x)|=kp$ for some integer $k>1$, and $kp\mid pq$; so, $k\mid q$ and hence ($k>1$) $k=q$. Therefore $|C_G(x)|=pq$, namely $C_G(x)=G$: contradiction. Likewise if we assume $|Z(G)|=q$.

If $Z(G)$ is trivial, then every nontrivial element of $G$ has centralizer of order $p$ or $q$. Therefore, the Class Equation yields: $$pq=1+k_pp+k_qq \tag 1$$ where $k_i$ is the number of conjugacy classes of size $i=p,q$. Since $\langle x\rangle=C_G(x)$ for every $x\in G\setminus\{e\}$, there are exactly $k_qq$ elements of order $p$ (they are the ones in the conjugacy classes of size $q$); but each subgroup of order $p$ contributes $p-1$ elements of order $p$, and two subgroups of order $p$ intersect trivially, then $k_qq=m(p-1)$ for some positive integer $m$ such that $q\mid m$ (because $q\nmid p-1$). Therefore, $(1)$ yields: $$pq=1+k_pp+m'q(p-1) \tag 2$$ for some positive integer $m'$; but then $q\mid 1+k_pp$, namely $1+k_pp=nq$ for some positive integer $n$, which replaced in $(2)$ yields: $$p=n+m'(p-1) \tag 3$$ In order for $m'$ to be a positive integer, necessarily $n=1$ (which in turn implies $m'=1$, though this is not relevant here). So, $1+k_pp=q$: but this is a contradiction if we assume $p\nmid q-1$. So, in that case, we are left with $Z(G)=G$, namely $G$ Abelian.

Now, $G$ has at most one subgroup of order $p$, because if they were two, say $H$ and $K$, then $HK$ would be a subgroup of $G$ of order $p^2$ (contradiction, as $p^2\nmid pq$). Likewise, $G$ has at most one subgroup of order $q$, because if they were two, say $H'$ and $K'$, then $H'K'$ would be a subgroup of $G$ of order $q^2$ (contradiction, as $q^2\nmid pq$). But then, there are at least $pq-1-(p-1)-(q-1)=(p-1)(q-1)\ge 2$ elements of order $pq$, and hence $G$ is cyclic.

Answer 4

Claim: Let $p,q$ be two primes such that $p>q$. And let $|G|=pq$. Then $G$ is solvable. Further if $q \neq 1 (\text{mod } p)$, then $G \cong \mathbb{Z}_{pq}$.

Proof: Using Sylow 3rd, $n_q | |G| = pq$ and $n_q =1 (\text{mod } p)$, which forces $n_q=1$, since $p<q$. So $G$'s Sylow $q$-subgroup $Q$ is normal in $G$. That further implies $\{e\}<Q \leq G$ is a solvable series.

Suppose further $q \neq 1 (\text{mod } p)$. Then $n_p=1$ must follow (By Sylow 3rd). Let $P,Q$ be the Sylow $p-,q-$ subgroup. Then $G \cong P \times Q$ must follow.