# counter example needed: existence of an isomorphism which maps a "diagonal" (non product) subgroup of a finite abelian group to a product subgroup.

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.

• How should $K$ not be of the form every finite abelian group is?? Apr 23 '21 at 12:15
• $K$ 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. Apr 23 '21 at 12:29
• 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. Apr 23 '21 at 12:49
• @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. Apr 23 '21 at 13:14
• @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. Apr 23 '21 at 16:36

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 $$8$$.
• It's the commutator $x^{-1}y^{-1}xy$. I have rewritten the presentation, and also added some explanation. Apr 23 '21 at 18:28
• "𝑥4 is the only nontrivial fourth power in 𝐺": $x^2$ ? "𝐾 contains 𝑥2": why ? Apr 23 '21 at 18:59
• 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. Apr 23 '21 at 19:05
• 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 ? Apr 23 '21 at 19:42