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

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.

• $|G| = 7\times 11$ and $7 \not\mid (11-1)$ so $G \cong \mathbb{Z}_{77}$. Commented May 27, 2017 at 13:49
• It's hard to answer this question without knowing how much you know. If you know Sylow's Theorems then it's easy, but you perhaps you don't. Do you know about conjugacy classes and centralizers? Commented May 27, 2017 at 15:45

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 $$p \not\mid (q - 1)$$.

• Didn't take neither Sylow groups nor $Aut(G)$ :)
– Nour
Commented May 27, 2017 at 14:07
• Why $7 \nmid (11-1)$ implies $\langle \sigma \rangle$ and $\langle \tau \rangle$ are normal? Does it just follow from the fact that number of syllow 7 and syllow 11 is 1? Commented Nov 16, 2018 at 0:00
• Yes. More generally, for any group $G$ of order $p q$ for primes $p < q$, Sylow forces any subgroup $\langle \tau \rangle \cong \Bbb Z_q$ to be normal. Since $p, q$ are coprime, $G = \Bbb Z_q \rightthreetimes_{\phi} \Bbb Z_p$ for any subgroup $\Bbb Z_p = \langle \sigma \rangle$ and for some homomorphism $\phi : \Bbb Z_p \to \operatorname{Aut}(\Bbb Z_q) \cong \Bbb Z_{q - 1}$. In particular, if $p \not\mid (q - 1)$, $\phi$ is trivial and so $G = \Bbb Z_p \times \Bbb Z_q \cong \Bbb Z_{pq}$. Commented Nov 16, 2018 at 4:02
• @Nour, if you haven't got Sylow at hands you should have stated it upfront in the question. Commented Dec 29, 2022 at 5:08

$$|G|$$=$$77$$ 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 $$n_{11}\mid77$$

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, $$H\congZ_{11}$$ so it is a cyclic group of order 11 and $$K\congZ_{7}$$ 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,$$G\congZ_{77}$$ and so it is a cyclic group of order 77

• From the comments of the top-scored answer we get that the OP hasn't got Sylow at hands. Commented Dec 29, 2022 at 4:43
• I just realized it
– user1134770
Commented Dec 29, 2022 at 4:49

An argument without Sylow and Cauchy goes as follows.

For $$G$$ a group of order $$pq$$, with $$p,q$$ primes (WLOG $$p), the center $$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); 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.

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 $$p. So $$G$$'s Sylow $$q$$-subgroup $$Q$$ is normal in $$G$$. That further implies $$\{e\} 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.