Let $G,G'$ be finite groups of orders $m,n$ respectively. What is the order of $G×G'$?

Question

Let $G,G'$ be finite groups of orders $m,n$ respectively. What is the order of $G×G'$?

I have started studying serge lang's undergraduate algebra. This is the question from books group theory exercise. Now lang defined $G×G'$ as follows:

Let $G,G'$ be groups. Let $G×G'$ be the set consisting of all pairs $(x,x')$ with $x\in G$and $x'\in G'$. If $(x,x') $and $(y,y') $are such paurs, define their product to be $(xy,x'y')$. Then $G×G'$ is a group.

Now according to the question, $G$ contains $m$ elements,and $G'$ has $n$ elements so $G×G'$ contains maximum $mn$ elements. but What is correct answer? Any help would be appreciated.

Answer 1

The group structures of $G$ and $G'$ only come into play when we want to define the group structure of $G \times G'$. For the number of elements of $G\times G'$, the group structure doesn't matter. All we need to know is what $G\times G'$ is as a set, and that's:

the set consisting of all pairs $(x,x')$ with $x\in G$ and $x' \in G'$.

There are $mn$ such pairs, simply because there's $m$ ways to choose $x$ and $n$ ways to choose $x'$.

Answer 2

You reasoning is almost correct. The direct product group has $mn$ elements (not maximum $mn$ elements). The notion of Cartesian product might help formalize it https://en.m.wikipedia.org/wiki/Cartesian_product. Just recall that groups are sets endowed with an operation.

Answer 3

If $G$ and $G'$ are cyclic groups of orders $m$ and $n$, where $m$ and $n$ are co-prime, then it can be shown that their product, $G \times G'$ is a cyclic group of order $m n$.

Let $(x, x') \in (G, G')$ be arbitrary. Then there exist integers $k$ and $l$ such that $x = g^k$ and $x' = (g')^l$, where $g$ is the generator of $G$ and $g'$ is the generator of $G'$. Since $m$ and $n$ are co-prime, there exist integers $a, b \in \mathbb{Z}$ such that $a n + b m = k - l$. We may rewrite this as $-b m + k = a n + l = N$ (say). Now,

$$ (g, g')^N = \left( g^{-b m + k}, (g')^{a n + 1} \right) = \left( g^{k}, (g')^l \right) = (x, x').$$

This proves that $(g, g')$ generates $G \times G'$.