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Let there $K\leq G \times H$ Then $K$ is of the form $G'\times H'$ where $G'\leq G$ And $H'\leq H$

Intuitively the definition of $G\times H$ a cartesian product of two groups, so there can not be an element in it, which is not a product of two subgroups.

By contradiction , Let assume that there is $(k_1,k_2)\in (A\times B)$ and $(k_3,k_4)\in (C\times D)$ Where $A,B,C,D$ are not a subgroup of $G$ or $H$ and not of each other, then $$(k_1,k_2)(k_3,k_4)=(k_1k_2,k_3k_4)\in K$$ but $k_1k_2\notin G$ and $k_3k_4\notin H$ contradiction

Is it correct?

newhere
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    Try to avoid using $G'$ as an arbitrary subgroup of $G$ because it stands for the commutator subgroup of $G$. – Wang Kah Lun Feb 24 '20 at 09:40
  • @AlanWang I am afraid I disagree with that. I find $G'$ far too useful to be reserved for the commutator subgroups, which I always denote by $[G,G]$ (or $G^{(1)}$ in the derived series). – Derek Holt Feb 24 '20 at 11:50

2 Answers2

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Consider $G$ and $H$ be vector spaces of dimension $2$ (over a finite field with their structure of additive group) and $K$ a subspace of dimension $3$ of $G\times H$.

Tsemo Aristide
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  • How can addition of two spaces of dimension 2, create a subspace of dimension 3? – newhere Feb 24 '20 at 09:36
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    The dimension of $G\times H$ is $dim(G)\times dim H$. You can take a subspace of a basis of $G\times H$ which contains $3$ elements to generate such a subspace. – Tsemo Aristide Feb 24 '20 at 09:38
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Hint(Counterexample):

Consider $G=H=\Bbb{Z}_2$ and $K=\{(0,0),(1,1)\}$.

Wang Kah Lun
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