Q:

Is $48$ the order of $S_3 \oplus \mathbb{Z}_8$?

I want to know if the following procedure is correct:
$G = S_3 \oplus \mathbb{Z}_8$
$S_3 = \{P_0, P_1, P_2, \mu_1, \mu_2, \mu_3\} $
$\mid S_3 \mid = 6 $
$\mathbb{Z}_8 = \{0, 1, 2, 3, 4, 5, 6, 7\} $
$\mid \mathbb{Z}_8 \mid = 8 $
Then $\mid G \mid = 48$

A:

You are correct.  In general, the order of $\displaystyle \bigoplus_{k=1}^n G_k$ where each $G_k$ is a finite group is $\displaystyle \prod_{k=1}^n |G_k|$.
Proof: We can proceed via strong induction.  For the trivial direct sum, the order of a single group $G_1$ is $|G_1|$, so we have the base case.  
Next, assume that the order of $\displaystyle \bigoplus_{k=1}^m G_k$ is $\displaystyle \prod_{k=1}^m |G_k|$ for all $1 < m \leq n$.  Now let's try to find the order of $\displaystyle \bigoplus_{k=1}^{n+1} G_k$.  We can rewrite this as:
$$\bigoplus_{k=1}^{n+1} G_k \cong \left( \bigoplus_{k=1}^n G_k \right) \oplus G_{n+1}$$
Now apply the induction hypothesis.  This is a direct sum of two groups whose orders are known, so its order is the product of the orders of the two groups themselves.  From this, we can conclude that the order of the group in question is indeed the product of the orders of all the $G_k$'s.

A (much shorter) combinatorial argument works as well ;)

