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While studying abstract algebra I encountered the following question:

Given an abelian group $G$ and its two sub-groups $H_1, H_2$, such that $H_1 \cap H_2=\{e\}$, prove that the group generated by $\langle H_1 \cup H_2 \rangle$ is isomorphic with the group $H_1 \times H_2$ (their Cartesian product, with the operation being the one from $G$ along the coordinates, so $(a,b)(c,d)=(ac, bd)$).

From what I understand I should suggest a bijection that preserves the operations of these groups. However I don't quite know how to go about this. I was thinking of the following: for each element $a$ of $\langle H_1 \cup H_2 \rangle$ we first express is at a linear combination of elements from $H_1 \cup H_2$ (so $a = (H_1)_1^{k_1}(H_1)_2^{k_2}...(H_2)_1^{q_1}(H_2)_2^{q_2}...$), separate the elements from $H_1$ and $H_2$ into two expressions ($R_1 = (H_1)_1^{k_1}(H_1)_2^{k_2}..., R_2=(H_2)_1^{q_1}(H_2)_2^{q_2}...$), and place the results of these expressions into their respective coordinates of the pair from the Cartesian product group (so $(R_1, R_2)$). Since $G$ is abelian, we don't have to worry about the order of the operations.

Does such proposition hold any water?

Anne Bauval
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Cezary Czubała
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  • You need to prove this is a homomorphism and that this homomorphism is invertible – FShrike Aug 31 '22 at 17:09
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    What you have shown is that the map $H_1\times H_2\to\langle H_1\cup H_2\rangle,;(a,b)\mapsto ab$ is onto. It is a morphism because $G$ is abelian. Now, compute its kernel ;-) – Anne Bauval Aug 31 '22 at 17:33

2 Answers2

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Lemma: Let $G$ be a group and $H\le G, K\triangleleft G$ then $H\vee K=HK$

Proof:

  1. $HK\subset H\vee K$.
  2. $HK\le G$.
  3. $H\vee K$ is the smallest subgroup of $G$ containing both $H$ and $K$. $\square$

Given $G$ is abelian, hence $H\vee K=HK$.

Verify (or see here):

$$\begin{align} \varphi : H\times K&\to HK,\\ (h, k) &\mapsto hk \end{align}$$

is an isomorphism.

Shaun
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SoG
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One way to define the (internal) direct product $G=H\bowtie K$ of two subgroups $H,K$ of a group is that all of the following is satisfied:

  • $G=HK$.
  • $H\cap K=\{e\}.$
  • $H,K\unlhd G$.

It is a standard exercise to show that $H\times K\cong H\bowtie K.$ (For a proof, see Theorem 9.6 of Gallian's, "Contemporary Abstract Algebra (Eighth Edition)" or see this question.)

Let $T=\langle H_1\cup H_2\rangle$.

We have that $H_i=\langle H_i\cup \{e\}\rangle$ is a subgroup of $T$. (As $T$ is the smallest subgroup of the ambient group that contains $H_i$. But $H_i$ is itself a group and so must be a subgroup of $T$.) Since the ambient group is abelian, $T$ is abelian, so each of the subgroups in $T$ is normal; hence $H_i\unlhd T$.

We are given that $H_1\cap H_2=\{ e\}$.

Since, again, $T$ is abelian, we have $T=H_1H_2$ (because any set product $AB$ is a subgroup iff $AB=BA$; see here).

Therefore,

$$\begin{align} T&\cong H_1 \bowtie H_2\\ &\cong H_1\times H_2. \end{align}$$

Shaun
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  • This answer doesn't really address the OP's question. You could just as well do a one-liner by saying that in the category Ab, the product and coproduct of two groups are the same but what use would that be to the OP? – Matematleta Aug 31 '22 at 18:01
  • It answers the titular question, @Matematleta, and, no, there is nothing stopping you from doing that one-liner. – Shaun Aug 31 '22 at 18:05
  • @Shaun: you could also replace "We have that Hi=⟨Hi∪{e}⟩ is a subgroup of T. (Use the one-step subgroup test.)" by "Hi is a subgroup of T". And I agree that "this answer doesn't really address the question", because it leaves as "a standard exercise" the proof that $H\times K\cong H\bowtie K$. – Anne Bauval Aug 31 '22 at 18:05
  • I have added a citation, @AnneBauval. But I disagree: we are not given that $H_i\le T$. It needs to be shown. – Shaun Aug 31 '22 at 18:11