Group Theory - Cartesian Product Given $G = C_3 \times C_2 \times C_4 \times C_5$ where $C_2 = \langle\{0,1\}, +2\rangle$; $C_3 = \langle\{0, 1, 2\}, +3\rangle$; and so on. Consider the group $(G, *)$, here the group operator $*$ is defined by the constituent groups in the Cartesian product. 
i) Write an expression for $t \in G$ as a vector and show where the components of vector $t$ are drawn from? 
ii) For the vectors $t, u \in G$, Compute $t*u = w \in G$ by showing the components of $w$. 
iii) What is the group identity $e_G$ ? Choose any $t \in G$, $t \neq e_G$ and compute $t^{-1}$ 
iv) List all the cosets of $G$ relative to the subgroup $C_3\times C_2 \times C_4$
Here's a picture for better formatting: http://imgur.com/ebgiHta
I've been attempting to gather my thoughts on this one but I'm struggling.
I've been thinking:
$t$ in $G$ is of the form $(T\bmod{3}, T \bmod 2, T \bmod 4, T \bmod 5)$ where $T$ is an integer.
$t*u = (T+U \bmod 3, T+U \bmod 2, T+U \bmod 4, T+U \bmod 5)$ where $t = (T \bmod 3, T \bmod 2, T \bmod 4, T \bmod 5)$ for some integer $T$, and similarly for $u$ and $U$.
$(0,0,0,0)$ is the identity
$t-1$ = $(3-T \bmod 3, 2-T \bmod 2, 4-T \bmod 4, 5-T \bmod 5)$ where $t = (T \bmod 3, T \bmod 2, T \bmod 4, T \bmod 5)$ for some integer $T$.
Any help would be awesome!
 A: A cartesian product $X \times Y$ consists of all ordered pairs $(x,y)$, where $x$ is in $X$ and $y \in Y$.  So $x$ and $y$ don't have to be the same in any way.  Similarly, $G = C_3 \times C_2 \times C_4 \times C_5$ should consist of all ordered quadruples $(t_1, t_2, t_3, t_4)$, where $t_1 \in C_3, t_2 \in C_2, t_3 \in C_4$, and $t_4 \in C_5$.  Your answer seemed to suggest that an element should have all the components be the same integer, which need not be the case.  
Yes, if $t = (t_1, t_2, t_3,t_4)$, and $u = (u_1, u_2, u_3, u_4)$, then $t \ast u$ should be $(t_1 + u_1, t_2 + u_2, t_3 + u_3, t_4 + u_4)$, and $(0,0,0,0)$ is the identity of $G$.
The problem statement is committing a slight abuse of notation, because $C_3 \times C_2 \times C_4$ is not really a subgroup of $G$ (it is literally not even a subset).  What they really mean is the subgroup $H:= C_3 \times C_2 \times C_4 \times \{0\}$, where by $\{0\}$ I mean the subset of $C_5$ whose only element is the zero element of $C_5$ (that is, $0 \pmod{5}$).
A trick with cosets: since $H$ is a group with $3 \cdot 2 \cdot 4 = 24$ elements, and $G$ is a group with $3 \cdot 2 \cdot 4 \cdot 5 = 120$ elements, a complete set of coset representatives for $H$ in $G$ should have exactly $\frac{|G|}{|H|} = \frac{120}{24} = 5$ elements.  So what you are looking for here are $5$ distinct elements $g_1, ... , g_5$ of $G$ such that $g_i \ast g_j^{-1}$ is not a member of $H$ for any $i \neq j$.  There is more than correct answer for a complete set of coset representatives, but there is a somewhat obvious choice you can make.
Let me know if you have any questions about anything I said.
