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Find the Subgroup of $\mathbb Z_4 \times \mathbb Z_2$ that is not the form of $ H \times K$, where $H$ is a subgroup of $\mathbb Z_4$ and $ K$ is a subgroup of $\mathbb Z_2$

Order elements of $\mathbb Z_4 \times Z_2$ are

  • $|(0 , 1)|= 2 ,|(2 ,0)|= 2,|(2 ,1)| = 2, |(3,0)| = 4, |(3,1)| = 4 , |(1,0)| = 4 ,|(1 , 1)| = 4 $

So by Lagrange`s Theorem possible order of a non trivial subgroups are $2 ,4$

Thus all the subgroups of order $2$ are

  • $H_1 = \{ (0,0) ,((0,1) \} , H_2 = \{ (0,0) , (2 , 0) \} , H_3 = \{ (0 ,0) , (2 ,1) \} $ ,

Thus all the subgroups of order $4$ are

  • $K_1 = < (3,0) > = \{(0,0) , (3,0), (2,0) , (1,0)\} , K_2 = < (3,1) > , K_3 = < (1 ,0)> , K_4 = <(1 ,1) > , K_4 = \{ (0 ,0) ,((0,1) , (2 ,0) , (2 ,1) \}$

I think there are no non trivial subgroup other than these Subgroups.

I think there is no Subgroup of this form of $ H \times K$, where $H$ is a subgroup of $\mathbb Z_4$ and $ K$ is a subgroup of $\mathbb Z_2$.

Is there exist a group $G_1 \times G_2$ such that $G_1 \times G_2$ has a subgroup of the form $ H \times K$, where $H$ is a subgroup of $G_1$ and $ K$ is a subgroup of $G_2$

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  • $\begingroup$ What is the question here? $K_3$ is $\mathbb{Z}_4\times \{0\}$, i.e. is of that form, while $K_4$ is not of that form. $\endgroup$ – vadim123 May 26 '15 at 16:20
  • $\begingroup$ @ Vadim : First one is- Find the Subgroup of $\mathbb Z_4 \times \mathbb Z_2$ of the form of $ H \times K$, where $H$ is a subgroup of $\mathbb Z_4$ and $\mathbb K$ is a subgroup of $\mathbb Z_2$ and second is - Is there exist a group $G_1×G_2$ such that$ G_1×G_2$ has a subgroup of the form$ H×K$, where H is a subgroup of $G_1$ and K is a subgroup of$ G_2$ $\endgroup$ – user120386 May 26 '15 at 16:28
  • $\begingroup$ Okay, what about the first sentence of the OP? It is neither of these. $\endgroup$ – vadim123 May 26 '15 at 16:29
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Hint:

Suppose $\; C_4=\langle a\rangle\;,\;\; C_2=\langle b\rangle\;$ (both written multiplicatively: cyclic groups of order $\;4,\,2\;$ , resp.) ,and look at the subgroup $\;\langle (a,b)\rangle\le C_4\times C_2\;$

If you insist in your notation (additive modulo), take the subgroup $\;\langle (1_4,1_2)\rangle\;$

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  • $\begingroup$ @ Timbuc :$< (a , b)> = \{ (e_1 , e_2) , (a,b) , (a^2 ,e_2) ,(a^3 ,b) \}$ = $H \times K$ , where $H = \{ e_1 ,a, a^2, a^3 \}$ and $K = \{e_2 , b\}$ $\endgroup$ – user120386 May 26 '15 at 16:39
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    $\begingroup$ @user120386 No, that is incorrect (those two do not have the same order). $\endgroup$ – Tobias Kildetoft May 26 '15 at 18:12
  • $\begingroup$ @User You've been addressed already by Tobias, but observe that with my (multiplicative) notation: $$L:=\langle (a,b)\rangle=\left\{\,(a,b),\,(a^2,1),\,(a^3,b),\,(a^4=1,1),\right\}$$ and also $$L\cap(C_4\times\{1\})=C_4\times\{1\}\;,\;\;L\cap(\{1\}\times C_2)=\{1\}\times C_2$$ which proves $\;L\;$ cannot be of the form $\;H\times K\;,\;\;H\le C_4\;,\;\;K\le C_2\;$ $\endgroup$ – Timbuc May 26 '15 at 18:54
  • $\begingroup$ @ Timbuc : Thanks,I was thinking in a wrong direction. $\endgroup$ – user120386 May 27 '15 at 1:58
  • $\begingroup$ The usage of "the" in the question is an implicit assumption that the answer is unique, but in fact there are two such subgroups, one of order 4 and one of order 2. The first one given by Timbuc, and the other one generated by $(a^2,b)$. $\endgroup$ – verret May 27 '15 at 9:51

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