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Let $G$ be a group, $H$ be a subgroup of $G$, and $N$ be a normal subgroup of $G$.

Consider the quotient map (or "natural projection homomorphism")

$$ \pi: G \to G/N $$ defined by $g \mapsto gN$, for all $g \in G$.

I know that $\pi(H)$ is a subgroup of $G/N$, but I occasionally see the assertion made that $\pi(H)$ is actually a subgroup of $H$, as well. However, I can't seem to prove (or disprove) this myself (or even find a proof anywhere), which makes me think it's either trivially true or not true (or, perhaps, very hard to prove).

So, basically, is $\pi(H)$ indeed a subgroup of $H$?

user1729
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thisisourconcerndude
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    Strictly speaking, $\pi(H)$ cannot be a subgroup of $H$ because it is not even a subset of $H$. The elements of $\pi(H)$ are cosets of $N$, of the form $hN, h \in H$ while elements of $H$ are just, well, elements $h \in H$. – Tob Ernack Dec 21 '17 at 00:28
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    Another question could be whether $\pi(H)$ is isomorphic to a subgroup of $H$. I am not sure this is true but haven't been able to find a proof or counterexample yet. – Tob Ernack Dec 21 '17 at 00:43
  • @Tob Ernack Actually, since this statement usually appears in divisibility arguments (in which case, I'm likely misstating it here), if $\pi(H)$ were simply isomorphic to a subgroup of $H$, that would make sense and be helpful. My suspicion is something like the quotient $H/(H \cap N)$... – thisisourconcerndude Dec 21 '17 at 00:48
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    Aha found another stackexchange question providing a counterexample to this: https://math.stackexchange.com/questions/909224/does-g-always-have-a-subgroup-isomorphic-to-g-n

    (This treats the special case when $G = H$). I have to admit I am disappointed I did not think of Lubin's counterexample which is quite neat.

     – Tob Ernack Dec 21 '17 at 01:24
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    I am skeptical you "occasionally see the assertion made that $\pi(H)$ is actually a subgroup of $H$," as that makes no sense. The image of $H$ is the subgroup $HN/N$ of $G/N$, which is isomorphic to $H/(H\cap N)$. So, the image is isomorphic to a quotient of $H$. (Even without $N$ being normal, $HN/N$ is a well-defined coset space and is isomorphic to $H/(H\cap N)$ as a left $H$-set. If you're interested in group actions.) Indeed, one of the isomorphism theorems says all homomorphic images of a group are isomorphic to a quotient of the group (in this case the group being $H$). – anon Dec 21 '17 at 05:37

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In general, using your notation, it is not true that $\pi(H)$ is a subgroup of $G$. For example, consider $G=(\mathbb{Z}, +)$ the integers under addition. This is the infinite cyclic group, and all its non-trivial subgroups are also infinite cyclic. Let $N=6\mathbb{Z}$ and $H=2\mathbb{Z}$. Then $\pi(H)$ is cyclic of order $3$, and therefore in this case $\pi(H)$ cannot be a subgroup of $G/H$. Note also that $G/N$ is not isomorphic to any subgroup of $G$.

You might want to look up semidirect products. A group $G$ is a semidirect product $N\rtimes K$ if $G/N$ "is a subgroup" of $G$, in a very specific sense.

user1729
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