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Let $H$ and $K$ be normal subgroups of $G$ with a trivial intersection.
$H \cap K = \{1\}$. Where $1$ is the unit.
Show that $hk = kh$ for every $k \in K, h \in H$.
Additionally, it asks if the statement still holds if only one of $K$ and $H$ is a normal subgroup.

Firstly, does $hk = kh$ for every $k \in K, h \in H$ mean $HK = KH$? It means that every element from $H$ commutes with every element from $K$. But does these two things tell the same thing?

What I did:
Since $ H \triangleleft G$
Let $h \in H$
for every $g \in G: ghg^{-1} \in H$. Similarly, $k \in K \ \forall g \in G : gkg^{-1} \in K$

It is also true that $khk^{-1} \in H$ and $hkh^{-1} \in K$.
So there exist such $h' \in H$ and $k' \in K$ that $khk^{-1} = h'$ and $hkh^{-1} = k'$.

$kh = h'k$
$hk = k'h$.

I do not know how to use that the intersection is trivial nor how to continue as I can't get anything out of this.

user26857
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Coupeau
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    For your "firstly" paragraph: $hk=kh:\forall:h\in H, k\in K\Rightarrow HK=KH$, but the converse is not true. For example, it is not true in $S_3$ when $H$ is generated by $(12)$ and $K$ is generated by $(123)$. Here, $K$ is normal in $S_3$ but $H$ is not, and more generally the reverse implication is not true when precisely one of the subgroups is normal, which the questions hints at a bit. – user1729 Jun 11 '19 at 10:42
  • Older questions about the same problem: https://math.stackexchange.com/q/147530 https://math.stackexchange.com/q/253131 https://math.stackexchange.com/q/773405 – Martin Sleziak Jun 11 '19 at 14:12
  • I've closed this question as a duplicate of an older question, as the OP has accepted an answer to the question addressed in the older questions. – user1729 Jun 11 '19 at 14:24
  • (The other question is addressed in my comment above.) – user1729 Jun 11 '19 at 14:25

1 Answers1

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Let $h \in H$ and $k \in K$. Then, $(hkh^{-1})k^{-1}=h(kh^{-1}k^{-1}) \in H\cap K$ (think about the parentheses and use that both $H$ and $K$ are normal subgroups). Since the intersection is trivial, the elements commute.

Tim
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    To complement they commute because $hkh^{-1}k^{-1}=1$ since $1$ is the only element in $H\cap K$ – kingW3 Jun 11 '19 at 12:39