Order of a direct product of finitely many groups equal to the product of the orders of the groups

Question

In order to further along a different proof, I need the following result:

Let $G_{1}, G_{2},\cdots , G_{k}$ be groups of orders $n_{1}$, $n_{2}$, $\cdots$, $n_{k}$, respectively. Then $\vert G_{1} \times G_{2} \times \cdots G_{k} \vert = \vert G_{1} \vert \cdot \vert G_{2} \vert \cdot \, \cdots \, \cdot \vert G_{k} \vert,\, \forall k \in \mathbb{N}$.

I was able to prove it, and I just wanted to make sure that my proof was correct.

This is my attempt:

For $\mathbf{k=1}$: $|G_{1}|+|G_{1}|$ (trivial).

For $\mathbf{k=2}$: Let $1 \leq a \leq n_{1} = |G_{1}|$ and $1 \leq b \leq n_{2}$. Then, $\forall x_{1,a}\in G_{1}$, there are $n_{2}$ possible $x_{2,b} \in G_{2}$ to pair together in $(x_{1,a},x_{2,b}) \in G_{1} \times G_{2}$. There are $n_{1}$ possible $x_{1,a}\in G_{1}$. Therefore, there are $n_{1}\cdot n_{2}$ elements $(x_{1,a},x_{2,b}) \in G_{1} \times G_{2}$, which implies that $|G_{1} \times G_{2}| = |G_{1}|\cdot |G_{2}|$.

Suppose true for $\mathbf{k = \tilde{m}}$: $|G_{1} \times G_{2} \times \cdots \times G_{\tilde{m}}| = |G_{1}|\cdot |G_{2}| \cdot \, \cdots \, \cdot |G_{\tilde{m}}| $.

Show true for $\mathbf{k = \tilde{m}+1}$: Let $1 \leq a \leq n_{1} = |G_{1}|$, $1 \leq b \leq n_{2} = |G_{2}|$, $\cdots$, $1 \leq m \leq n_{\tilde{m}} = |G_{\tilde{m}}|$, $1 \leq n \leq n_{\tilde{m}+1} = |G_{\tilde{m}+1}|$.

For all $x_{1,a}\in G_{1}$, $x_{2,b}\in G_{2}$, $\cdots$, $x_{\tilde{m},m} \in G_{\tilde{m}}$, $(x_{1,a}, x_{2,b}, \cdots , x_{\tilde{m},m})\in G_{1} \times G_{2} \times \cdots \times G_{\tilde{m}}$.

By the induction hypothesis, there are $n_{1}\cdot n_{2} \cdot\, \cdots \, \cdot n_{\tilde{m}}$ such tuples in $G_{1} \times G_{2} \times \cdots \times G_{\tilde{m}}$.

Let $1 \leq q \leq n_{1}n_{2}\cdots n_{\tilde{m}}$; also, let $w = (x_{1,a}, x_{2,b}, \cdots , x_{\tilde{m},m})$. If we fix $q$, then there are $n_{\tilde{m}+1}$ possible pairings $(w_{q},x_{\tilde{m}+1})$.

Since there are $n_{1}n_{2}\cdots n_{\tilde{m}}$ possible values of $q$, this means that there are $q \cdot n_{\tilde{m}+1} = (n_{1}n_{2}\cdots n_{\tilde{m}})\cdot n_{\tilde{m}+1}$ possible tuples $(x_{1,a},x_{2,b},\cdots , x_{\tilde{m},m},x_{\tilde{m}+1,m})$.

Hence $|G_{1} \times G_{2} \times \cdots \times G_{\tilde{m}} \times G_{\tilde{m}+1}| =\vert G_{1} \vert \cdot \vert G_{2} \vert \cdot \, \cdots \, \cdot \vert G_{\tilde{m}} \vert \cdot \vert G_{\tilde{m}+1} \vert$.

Thus, the equality holds $\forall k \in \mathbb{N}$.

Is this proof correct, and if not, what can I do to fix it?

Thank you.

Answer 1

Your proof is fine, but you can make the induction step a lot shorter. To wit, we can write $G_1 \times G_2 \times \cdots \times G_n \times G_{n+1} = H \times G_{n+1}$, where $H = G_1 \times G_2 \times \cdots \times G_n$. You can now immediately arrive at the conclusion you want since $\displaystyle |H| = \prod_{j=1}^n |G_n|$ by the induction hypothesis, and you previously took care of the $k=2$ case.

Answer 2

Your proof is fine, but it is unnecessary to do induction. This is a combinatorics/counting problem.

By definition, the elements of the group $G_1 \times \cdots \times G_K$ are precisely the tuples $(g_1,\ldots,g_K)$ where $g_i \in G_i$ for each $i=1,\ldots,K$. You just want to count how many such tuples there are. There are $|G_1|$ choices for $g_1$, $|G_2|$ choices for $g_2$, and so on, so $|G_1 \times \cdots \times G_K| = |G_1| \cdots |G_K|$.