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We define the following operation on the direct product set $G=\mathbb Z /(10\mathbb Z)\times\mathbb Z /(4\mathbb Z)$. For $(i_1,j_1),(i_2,j_2)\in \mathbb Z/(10\mathbb Z) \times \mathbb Z /(4\mathbb Z)$, $$(i_1,j_1)(i_2,j_2)=(i_1+3^{j_1}i_2, j_1+j_2). $$ We also define $3^{j_1}=3^k+10\mathbb Z \in \mathbb Z/(10\mathbb Z)$ if $j_1=k+4\mathbb Z \in \mathbb Z /(4\mathbb Z)$.

First, I can prove that $G$ together with the operation satisfies all of group axioms. Next my goal is to show that: $$G \cong \langle a,b~|~a^{10}=1,b^4=1,bab^{-1}=a^3 \rangle.$$ Let $x,y \in G$ such that $x=(1,0), y=(0,1)$. It's easy to check that $x^{10}=1,y^4=1,yxy^{-1}=x^3$. How can I have the conclusion? I think this somehow relates to the universal property of free groups. However I couldn't make it clear. Any help would be appreciated.

user
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1 Answers1

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You are almost done. What you have to prove that this $x=(1,0)$ and $y=(0,1)$ generate this group $G$ which is indeed true see $(i,j)=x^iy^j$.

So what you have is this a map $f:G \to <a,b |a^{10}=b^4=1;bab^{-1}=a^3>$ which sends generator of $G$ $x,y$ to generator of $<a,b|a^10=b^4=1;bab^{-1}=a^3>$ $a,b$. Then you are done with isomorphism.

Ri-Li
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  • Can you show the property of $f$ that for every $x_1, x_2 \in G$ we have $f(x_1x_2)=f(x_1)f(x_2)$? – user Sep 18 '15 at 12:00
  • Yes this follows from the generator relation. See carefully. – Ri-Li Sep 18 '15 at 12:02
  • For example, as above call $x, y$ are generators of $G$. By the property of your map $f$, we have $f(x)=a, f(y)=b$. How can you show that $f(xy)=ab$? – user Sep 18 '15 at 12:33
  • I am defining the map in that way such as $f(x)=a$, $f(y)=b$ and $f(x^i_1y^j_1x^i_2y^j_2)=a^i_1b^j_1a^i_2b^j_2$. Are you getting my point? – Ri-Li Sep 18 '15 at 16:20
  • $f(x^{i_1}y^{j_1}x^{i_2}y^{j_2}....)=a^{i_1}b^{j_1}a^{i_2}b^{j_2}...$ sorry. – Ri-Li Sep 18 '15 at 17:40
  • What is the typical form of elements in $G$? Isn't it be $x^iy^j$? Why $x^{i_1}y^{j_1}x^{i_2}y^{j_2}....$? – user Sep 19 '15 at 07:45
  • I have not used the relation $bab^{-1}$ there in the previous comment. I want to write it in the free group form. Such a way you understand the expression well. – Ri-Li Sep 19 '15 at 08:09
  • Just a quick comment. You define a map which sends elements of $G$ to elements of $H= \langle a,b~|~a^{10}=1,b^4=1,bab^{-1}=a^3 \rangle$. I know from the definition of free group that every element of $H$ can be written as a word forming from $a, b$. Meanwhile, the typical element of $G$ is of the form $x^iy^j$. Can you give the precise definition of your map $f$? – user Sep 19 '15 at 08:19
  • I suppose you are not getting the generator relation properly. Can you say how to show two cyclic group of same order are isomorphic?? – Ri-Li Sep 19 '15 at 12:53
  • It isn't a problem for two cyclic groups. If you need that I can give you https://proofwiki.org/wiki/Cyclic_Groups_of_Same_Order_are_Isomorphic – user Sep 19 '15 at 14:57
  • I do not need that. Probably you have to understand that properly. Then just follow the process with generators here also. Define the same way. I have also given you the process. – Ri-Li Sep 20 '15 at 16:41
  • There you have only one generator here you have two. Thats it. – Ri-Li Sep 20 '15 at 16:48
  • And after that if you still have a confusion then just check by hand that the relation in $5$ th comment give you a homomorphism. – Ri-Li Sep 20 '15 at 18:21