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I am working with a $pq$ group. Where $p=7,q=11$.

I have so far managed to prove that such a group is cyclic and hence abelian.

My question asks what is this group isomorphic to and I don't really know. I do remember learning that we can classify all finite abelian groups into direct products of cyclic groups but I don't remember the specifics sorry.

As a guess I would think maybe our group of order $77$ is either isomorphic to $\mathbb{Z}_{77}$ or maybe $\mathbb{Z}_{7} \times \mathbb{Z}_{11}$ but I really don't know how to show this?

Ben B
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  • Every cyclic group of order $n$ is isomorphic to $\mathbb{Z_{n}}$. – Jonathan Hebert Mar 16 '17 at 19:07
  • Is this person wrong then? http://mymathforum.com/abstract-algebra/5275-classify-groups-order-77-a.html – Ben B Mar 16 '17 at 19:10
  • $\mathbb{Z}{77}$ is isomorphic to $\mathbb{Z}_7\times\mathbb{Z}{11}$ via $1\mapsto(1,1)$ – Robert Chamberlain Mar 16 '17 at 19:22
  • No, they aren't, but they aren't fully using the information provided. Since $7$ and $11$ are coprime, $\mathbb{Z}{7} \times \mathbb{Z}{11}$ is isomorphic to $\mathbb {Z}{77}$. I think $\mathbb {Z}{77}$ is a better answer, though both are correct. – Jonathan Hebert Mar 16 '17 at 19:23
  • If $\gcd (m, n) = 1$, then $$\mathbb Z_m \times \mathbb Z_n \cong \mathbb Z_{mn}$$ In your case, since $\gcd(7, 11) = 1$, $$\mathbb Z_7\times \mathbb Z_{11} \cong \mathbb Z_{77}$$ – amWhy Mar 17 '17 at 01:08
  • How is this question be duplicate while the target question did not mention abelian? – tsh May 23 '25 at 09:59

2 Answers2

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I have so far managed to prove that such a group is cyclic

Cyclic groups are uniquely determined by their order. Every cyclic group is isomorphic either to $\mathbb{Z}$ or to $\mathbb{Z}_n$ for some $n$. This is quite trivial. If $G$ is a cyclic group generated by $g\in G$ then we have a group homomorphism

$$f:\mathbb{Z}\to G$$ $$f(n)=g^n$$

It is onto (because $g$ generates $G$) and thus by the first isomorphism theorem $G\simeq\mathbb{Z}/\ker(f)$. If $g$ is not of finite order then $\ker(f)=0$ and thus $G\simeq\mathbb{Z}$. Otherwise $|g|=n$ and $\ker(f)=n\mathbb{Z}$ and thus $G\simeq\mathbb{Z}_n$.

More generally by the fundamental theorem of finite abelian group, if $G$ is a finite abelian group of order $n=p_1^{\alpha_1}\cdots p_n^{\alpha_n}$ ($p_i$ prime) then

$$G\simeq A_{1}\oplus\cdots\oplus A_{n}$$ where each $A_i$ is a finite abelian group of order $p_i^{\alpha_i}$. Furthermore if $A$ is a finite abelian group of order $p^k$ then

$$A\simeq\mathbb{Z}_{p^{\beta_1}}\oplus\cdots\oplus\mathbb{Z}_{p^{\beta_m}}$$

where

$$\beta_1+\cdots +\beta_m=k$$

So all in all every finite abelian group is a product of of cyclic groups of order $p^k$ for some primes $p$. Note that

$$\mathbb{Z}_{p^a}\oplus\mathbb{Z}_{p^b}\simeq\mathbb{Z}_{p^{a+b}}$$

if and only if either $a=0$ or $b=0$.

In particular if $p, q$ are distinct primes and $G$ is abelian of order $|G|=pq$ then

$$G\simeq\mathbb{Z}_{pq}\simeq\mathbb{Z}_{p}\oplus\mathbb{Z}_{q}$$

freakish
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All finite groups of order $pq$ are classified, see here. If $q$ does not divide $p-1$ there is only one group of order $pq$, which is cyclic. Hence it is isomorphic to $$ C_p\times C_q\cong C_{pq}, $$ for distinct primes $p$ and $q$. The isomorphism follows from the CRT because $p$ and $q$ are corpime.

Our example, with $(p,q)=(7,11)$ does satisfy $p\nmid ( q-1)$.

Community
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Dietrich Burde
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