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Consider a finite abelian group:

$$G \cong \mathbb{Z}/p_1^{\alpha_1}\mathbb{Z} \times\mathbb{Z}/p_2^{\alpha_2}\mathbb{Z} \times\dots\times\mathbb{Z}/p_n^{\alpha_n}\mathbb{Z} $$

Let $K$ be a subgroup of $G$, which is not a product of cyclic group, i.e. not of the form: $$K = p_1^{\beta_1}\mathbb{Z}/p_1^{\alpha_1}\mathbb{Z} \times p_2^{\beta_2}\mathbb{Z}/p_2^{\alpha_2}\mathbb{Z} \times\dots\times p_n^{\beta_n}\mathbb{Z}/p_n^{\alpha_n}\mathbb{Z} $$ with $\beta_i \leqslant \alpha_i.$

I know the example of in $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z}$ of $K_1=\langle (0,2)\rangle$ and $K_2=\langle(1,0)\rangle $: $K_1 \cong K_2$ but there is no isomorphism of $G$ that maps $K_1$ to $K_2$. I would be inclined to extrapolate this and think that we cannot prove the existence of an isomorphism of $G$ that maps any subgroup $K$ which is not a product into a product subgroup. If I am right I would need a counter example to prove this non existence assertion. Please answer in this specific case of finite abelian group and for the mapping to "diagonal" to product subgroup.

edit for clarification: Let $$G =\mathbb{Z}/p_1^{\alpha_1}\mathbb{Z} \times\mathbb{Z}/p_2^{\alpha_2}\mathbb{Z} \times\dots\times\mathbb{Z}/p_n^{\alpha_n}\mathbb{Z} $$. Let $K$ be a subgroup of $G$ which is not a product of cyclic groups of prime power orders. Then there exist an isomorphism $\varphi$ of $G$ such that the image of $K$, $\varphi(K)$ is a product of cyclic groups of prime power orders.

I need a counter example to prove the assertion is false. but it might be true; in the latter case i need a full proof or a link to such proof.

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  • $\begingroup$ How should $K$ not be of the form every finite abelian group is?? $\endgroup$ Apr 23 '21 at 12:15
  • $\begingroup$ $K$ is isomorphic itself to a product of cyclic groups. Here I am discussing the existence of an isomorphism of $G$ which contains $K$ and whose restriction on$K$ sends $K$ to the desired form. Any subgroup of $G$ is not necessarily a product. $\endgroup$ Apr 23 '21 at 12:29
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    $\begingroup$ Does $V_4 = (\mathbb Z/2 \mathbb Z)^2$ answer your question? It has three subgroups of order $2$, one of which is "diagonal", and the automorphisms of $V_4$ give a cyclic permutation of those three subgroups. $\endgroup$ Apr 23 '21 at 12:49
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    $\begingroup$ @Pierre-PaulT. I am afraid that you are being very unclear indeed. For example, I have absolutely no idea what you mean in your previous comment by the sentence "Any subgroup of $G$ is not necessarily a product". I don't understand what you are asking, and I think it is up to you to make it clearer. You should avoid using the word "any", which is ambiguous. $\endgroup$
    – Derek Holt
    Apr 23 '21 at 13:14
  • $\begingroup$ @Mees de Vries: it does not. Let's try to rephrase it, since obviously it looks confusing (but implying it is totally incomprehensible is a tad exaggerated in my opinion. definition of any: whichever of a specified class might be chosen). please refer the edit of my original post. $\endgroup$ Apr 23 '21 at 16:36
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I think the subgroup $K=\langle x^2y \rangle$ of order $4$ in the group $G=\langle x,y \mid x^8=y^2=1, xy=yx\rangle$ is a counterexample.

To see that, note that $x^4$ is the only nontrivial fourth power in $G$, so all cyclic subgroups of order $8$ contain $x^4$. But $K$ contains $x^4 = (x^2y)^2$, so $K$ cannot be a direct factor of order $4$. Since $x^2y$ is not a square, it also cannot be contained in a direct factor of order $8$.

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  • $\begingroup$ what is [𝑥,𝑦] ? $\endgroup$ Apr 23 '21 at 18:06
  • $\begingroup$ It's the commutator $x^{-1}y^{-1}xy$. I have rewritten the presentation, and also added some explanation. $\endgroup$
    – Derek Holt
    Apr 23 '21 at 18:28
  • $\begingroup$ "𝑥4 is the only nontrivial fourth power in 𝐺": $x^2$ ? "𝐾 contains 𝑥2": why ? $\endgroup$ Apr 23 '21 at 18:59
  • $\begingroup$ Sorry, multiple typos. I keep getting confused by the fact that $x^2$ has order $4$ and $x^4$ has order $2$. I hope it's correct now. $\endgroup$
    – Derek Holt
    Apr 23 '21 at 19:05
  • $\begingroup$ ok good. One last question related to the first part of your justification: your $G$ is $G\cong\mathbb{Z}/\mathbb{8Z}\times\mathbb{Z}/\mathbb{2Z}$. This decomposition as product of cyclic groups of prime powers orders being unique, it was hopeless anyway to find an isomorphism of $G$ which sends $K$ to $\mathbb{Z}/\mathbb{4Z}$ ? Am I right ? $\endgroup$ Apr 23 '21 at 19:42

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