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Suppose $G_1=\langle X_1|R_1\rangle$, $G_2=\langle X_2|R_2\rangle$, $X_1\cap X_2=\emptyset$. I want to show that $G_1\times G_2=\langle X_1\cup X_2|R_1\cup R_2\cup[X_1,X_2] \rangle$.

By using universal property of free groups, I have obtained homomorphisms $f_i:F(X_i)\rightarrow F(X_1 \cup X_2)$. Also I can define homomorphisms $\pi_i:F(X_i) \rightarrow G_i$ such that $\ker(\pi_i)= R_i^{F(X_i)}=$ normal closure of $R_i$.

I want to define a group $G = \langle X_1\cup X_2|R\rangle$ with $R$ satisfying the conditions of direct product, and somehow $R$ contains $[X_1,X_2]$. But I'm kind of stuck here.

Shaun
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Sid Caroline
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    Try sending $(g,h)$ to $gh$ in the group defined using generators and relations – Exit path Jun 12 '17 at 11:24
  • Can you be a little more explicit? Suppose that I define $H=\langle X_1 \cup X_2 | R_1 \cup R_2 \cup [X_1,X_2]\rangle$, and $G_1 \times G_2={(g,h)|g\in G_1, h \in G_2}$, and I define a function $\phi:G_1\times G_2 \rightarrow H$, sending $(g,h)$ to $gh$, are you saying that I should try to prove $\phi$ is an isomorphism? – Sid Caroline Jun 12 '17 at 11:54
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    Yup. Try to show first that it's a homomorphism, then that it's injective, and last that it's surjective. At each step you'll have to use the relations in $H$. – Exit path Jun 12 '17 at 11:56
  • Not exactly sure how to prove injective and surjective. I feel like I need to use the 3 conditions of internal direct product somwhere. – Sid Caroline Jun 12 '17 at 12:18
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    You could define a homomorphism $\psi:H \to G_1 \times G_2$ that restricts to the identity on $X_1$ and $X_2$, and then show that $\phi$ and $\psi$ are mutually inverse maps. That implies that they are both bijective. – Derek Holt Jun 12 '17 at 16:43
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    If $gh=e$, then $g=h^{-1}$. What can you say about $g$ and $h$? Hint: They're words in $X_1$ and $X_2$ respectively. For surjectivity, take an arbitrary word in $X_1 \cup X_2$, and see if you can rewrite it in an appropriate sense. Or you could take Derek Holt's approach. Either way should work – Exit path Jun 12 '17 at 21:57
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    @leibnewtz I agree that surjectivity is easy, but I don't see how to use your hint to deduce injectivity. That's why I suggested defining the inverse homomorphism. – Derek Holt Jun 13 '17 at 07:53
  • @DerekHolt sir, if I define $(1,a)\mapsto 1$ and so on, then how to define rest of the elements? – MANI Nov 27 '19 at 18:34

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