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Where does the following argument break down?

Let $G$ be a group, with $H \triangleleft G$, $K \leq G$ and $H \cap K = 1$. Taking the semidirect product, we have $G \cong H \rtimes K$. Identifying $K$ with $1 \times K$, we consider conjugation of $K$ with any element in $H \rtimes K$. Pick any element $(h, k)$, then $(h, k)^{-1} \cdot (1, K) \cdot (h, k) = (1, K)$ Therefore, we conclude that $K$ must be normal in $G$ as well. But this is clearly wrong.

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It is false that $(h, k)^{-1} \cdot (1, K) \cdot (h, k) = (1, K)$. Note that




so that

$$(h,k)(1,a)(h,k)^{-1}=(h,k)(1,a)(k^{-1}h^{-1}k,k^{-1})=(h,ka)(k^{-1}h^{-1}k,k^{-1})$$ $$=(h(ka)(k^{-1}h^{-1}k)(ka)^{-1},kak^{-1})=(h(kak^{-1})h^{-1}(kak^{-1})^{-1},kak^{-1})$$

and there is no reason the first entry of this must be $1$ (unless $a=1$, of course).

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Recall that writing $G \simeq H \rtimes K$ means that $(h,k) \mapsto hk$ is a just bijection between $H \times K$ and $G$, but not necessarily a group morphism (in which case you would have a direct product).

In your case, you have

$(h,k_1)^{-1} . (1,k_2) . (h, k_1 ) = k_1^{-1} h^{-1} k_2 h k_1$

Which you can rewrite:

$(h,k_1)^{-1} . (1,k_2) . (h, k_1 ) = (h^{k_1})^{-1} k_2^{k_1} h^{k_1}$

And since $k_2^{k_1}$ is in $K$, you have

$ (h,k_1)^{-1} . (1,K) . (h, k_1 ) = (h^{k_1})^{-1} (1,K) h^{k_1}$

This means that conjugates of $(1,K)$ can always be written as conjugates by an element of $H$, but that's all you can say in general.

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