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I tried proving this in the following manner, but I am not confident with these types of problems so any verification would be appreciated. Thank you.

Let $A = \{(gH, gK): g \in G\}$

Define $\phi$ : $G$ $\rightarrow$ $A$ by $\phi(g)=(gH,gK)$

First we'll show $\phi$ is a homomorphism:

$\phi(gg')=(gg'H,gg'H)=(gHg'H,gHg'H)=(gH,gH)(g'H,g'H)=\phi(g)\phi(g')$

Now we'll show $\phi$ is bijective, and thus an isomorphism:

The codomain of $\phi$ is {$(gH,gK):g \in G$} so $\phi$ is clearly onto. Now suppose $g_1\not= g_2$ and $\phi(g_1)=\phi(g_2)$. Then $(g_1H,g_1K)=(g_2H,g_2K)$ $\Rightarrow$ $g_1H=g_2H$ and $g_1K=g_2K$ $\Rightarrow$ $g_2^{-1}g_1\in H$ and $g_2^{-1}g_1 \in K$, a contradiction since we assumed $H\bigcap K = \{e\}$

$\square$

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    $\begingroup$ Your argument is sound. You could also prove injectivity directly, instead of by contradiction. $\endgroup$ Oct 21 '18 at 6:43
  • $\begingroup$ Ah so you would just say $g_2^{-1}g_1 \not= e$ $\Rightarrow$ $g_2^{-1}g_1 \notin H \bigcap K$ $\Rightarrow$ either $g_1H \not= g_2H$ or $g_1K \not= g_2K$? $\endgroup$
    – PacoK
    Oct 21 '18 at 7:10
  • $\begingroup$ I would say $g_2^{-1}g_1 \in H \cap K = \{e \} \implies g_2^{-1} g_1 = e \implies g_1 = g_2$. $\endgroup$ Oct 21 '18 at 7:18
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You could make this cleaner by just directly calculating the kernel (though like somebody already said, your argument is valid):

\begin{align*} \ker\phi &=\{g\in G\mid (gH,gK)=(H,K)\}\\ &= \{g\in G\mid gH=H\text{ and }gK= K\}\\ &= \{g\in G\mid g\in H\text{ and }g\in K\}\\ &= H\cap K\\ &= \{e\}\end{align*}

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  • $\begingroup$ Oh, that makes a lot of sense. I tried going down that route at first, but for some reason I second guessed myself. $\endgroup$
    – PacoK
    Oct 21 '18 at 7:16

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