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It is natural. But I want to prove exactly.

My idea is to use isomorphism theorem.

Let $f: G \to G/H$ be the quotient map, then $f$ is a homomorphism and obviously surjective.

Hence, I just need to show $\ker(f) = K$.

However, I couldn't find the way. Definitely, $\ker(f) = H $ and $H \cong K$

But I can say $\ker (f) = K$.

How to prove it?

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This is not true. Consider $G=\mathbb{Z}_2\oplus\mathbb{Z}_4$, $H=\langle (1,0)\rangle$, and $K=\langle (0,2)\rangle$.

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  • 5
    $\begingroup$ Indeed, a classical example. Another one is $G = \mathbb{Z}, H = 2\mathbb{Z},K = 3\mathbb{Z}$. $\endgroup$ – Alex Wertheim Jun 7 '17 at 7:14
  • $\begingroup$ Thanks.. I regret my ignorance. $\endgroup$ – shashack Jun 7 '17 at 7:18
  • 1
    $\begingroup$ @shashack Don't fret, everyone that studies math has tried to prove the impossible. $\endgroup$ – Aweygan Jun 7 '17 at 7:23

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