I am asked which subgroups of $G = \mathbb{Z}_4 \times \mathbb{Z}_4$ have the same image in $G/G[2]$, where $G[2] = \{ g \in G: \operatorname{ord}(g) \,|\, 2\}$).
I have determined all subgroups and I know that $G/G[2]$ is the set of cosets $\{ g + G[2]: g\in G\}$. But I don't really understand what they mean by image.
I know that $(0,2) + G[2] = (2,0) + G[2] = (2,2) + G[2] = (0,0) + G[2] = G[2]$, so in this sense the subgroups $\langle (2,0) \rangle$, $\langle (2,0) \rangle$, $\langle (2,2)\rangle$ and $\langle (2,0),(0,2) \rangle$ can somehow be identified with $G[2]$. But all other subgroups contain elements from different cosets, so I'm a bit perplexed by what I am asked to do.
$G[2] = \{(0, 0), (0, 2), (2, 0), (2, 2)\}.$ Consider the natural homomorphism $\pi : G \rightarrow G/G[2].$ What are the images of the different subgroups of $G$ under the map $\pi$?