This is Prob. 2, Sec. 14, in the book A First Course In Abstract Algebra by John B. Fraleigh, 7th edition:
How many elements are there in the factor group $$ \left( \mathbb{Z}_4 \times \mathbb{Z}_{12} \right) / \big( \langle 2 \rangle \times \langle 2 \rangle \big)? $$
My Attempt:
We note that $\mathbb{Z}_4 \times \mathbb{Z}_{12}$ has $4 \times 12 = 48$ elements, and we also note that $$ \langle 2 \rangle \times \langle 2 \rangle = \{ \ (0, 0), (2, 2), (0, 4), (2, 6), (0, 8), (2, 10) \} $$ so that $\langle 2 \rangle \times \langle 2 \rangle$ has $6$ elements.
Therefore the factor group $\left( \mathbb{Z}_4 \times \mathbb{Z}_{12} \right) / \big( \langle 2 \rangle \times \langle 2 \rangle \big)$ has $48 / 6 = 8$ elements.
More precisely, the factor group $\left( \mathbb{Z}_4 \times \mathbb{Z}_{12} \right) / \big( \langle 2 \rangle \times \langle 2 \rangle \big)$ has the following left (or right) cosets: $$ \begin{align} \langle 2 \rangle \times \langle 2 \rangle \ &= \ (0,0) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= \ (2, 2) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= \ (0, 4) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= \ (2, 6) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= \ (0, 8) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= \ (2, 10) + \big( \langle 2 \rangle \times \langle 2 \rangle \big), \end{align} $$ $$ \begin{align} \{ \ (1,0), (3, 2), (1, 4), (3, 6), (1, 8), (3, 10) \ \} \ &= \ (1, 0) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= \ (1, 4) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= \ (1, 8) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= \ (3, 2) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= \ (3, 6) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= \ (3, 10) + \big( \langle 2 \rangle \times \langle 2 \rangle \big), \end{align} $$ $$ \begin{align} \{ \ (0, 1 ), (2, 3), (0, 5), (2, 7), (0, 9), (2, 11) \ \} \ &= \ (0, 1) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= \ (0, 5) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= \ (0, 9) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= \ (2, 3) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= \ (2, 7) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= \ (2, 11) + \big( \langle 2 \rangle \times \langle 2 \rangle \big), \end{align} $$ $$ \begin{align} \{ \ (1, 1), (3, 3), (1, 5), (3, 7), (1, 9), (3, 11) \ \} \ &= \ (1, 1) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= \ (1, 5) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= \ (1, 9) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= \ (3, 3) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= \ (3, 7) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= \ (3, 11) + \big( \langle 2 \rangle \times \langle 2 \rangle \big), \end{align} $$ $$ \begin{align} \{ \ (0, 2), (2, 4), (0, 6), (2, 8), (0, 10), (2, 0) \ \} \ &= \ (0, 2) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= \ (0, 6) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= \ (0, 10) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= \ (2, 0) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= \ (2, 4) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= \ (2, 8) + \big( \langle 2 \rangle \times \langle 2 \rangle \big), \end{align} $$ $$ \begin{align} \{ \ (0, 3), (2, 5), (0, 7), (2, 9), (0, 11), (2, 1) \ \} \ &= \ (0, 3) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= \ (0, 7) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= \ (0, 11) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= \ (2, 1) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= \ (2, 5) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= \ (2, 9) + \big( \langle 2 \rangle \times \langle 2 \rangle \big), \end{align} $$ $$ \begin{align} \{ \ (1, 2), (3, 4), (1, 6), (3, 8), (1, 10), (3, 0) \ \} \ &= \ (1, 2) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= (1, 6) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= (1, 10) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= (3, 0) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= (3, 4) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= (3, 8) + \big( \langle 2 \rangle \times \langle 2 \rangle \big), \end{align} $$ and $$ \begin{align} \{ \ (1, 3), (3, 5), (1, 7), (3, 9), (1, 11), (3, 1) \ \} \ &= \ (1, 3) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= (1, 7) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= (1, 11) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= (3, 1) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= (3, 5) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \\ &= (3, 9) + \big( \langle 2 \rangle \times \langle 2 \rangle \big) \end{align}. $$
Is my answer correct? Is my calculation correct in each and every detail? Is my logic correct?
If not, then where are the problems in my attempt?
$$|\mathbb{Z}_4 \times \mathbb{Z}_{12}| = 48 $$ $$|\langle 2 \rangle \times \langle 2 \rangle| = 2 \cdot 6 = 12$$ $$|\mathbb{Z}_4 \times \mathbb{Z}_{12} / \langle 2 \rangle \times \langle 2 \rangle| = 4$$
$$(0,0) +\langle 2 \rangle \times \langle 2 \rangle = \left\{ \begin{array}{c} (0,0) & (0,2) & (0,4) & (0,6) & (0,8) & (0,10)\\ (2,0) & (2,2) & (2,4) & (2,6) & (2,8) & (2,10)\\ \end{array}\right\}$$
$$(1,0) +\langle 2 \rangle \times \langle 2 \rangle = \left\{ \begin{array}{c} (1,0) & (1,2) & (1,4) & (1,6) & (1,8) & (1,10)\\ (3,0) & (3,2) & (3,4) & (3,6) & (3,8) & (3,10)\\ \end{array}\right\}$$
$$(0,1) +\langle 2 \rangle \times \langle 2 \rangle = \left\{ \begin{array}{c} (0,1) & (0,3) & (0,5) & (0,7) & (0,9) & (0,11)\\ (2,1) & (2,3) & (2,5) & (2,7) & (2,9) & (2,11)\\ \end{array}\right\}$$
$$(1,1) +\langle 2 \rangle \times \langle 2 \rangle = \left\{ \begin{array}{c} (1,1) & (1,3) & (1,5) & (1,7) & (1,9) & (1,11)\\ (3,1) & (3,3) & (3,5) & (3,7) & (3,9) & (3,11)\\ \end{array}\right\}$$