I was actually asking the same question in here. However, the give answer didn't satisfy me. For the reader's convenience I'll re-write the post as follows:

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 $x^{10}=1,y^4=1,yxy^{-1}=x^3$. By the previous post, I was advised to know that $x$ and $y$ generate this group $G$ which is indeed true as $G \ni (i,j)=x^iy^j$. I cannot think that I can stop here and get the desired conclusion. I found some useful arguments that may fulfill the question.

Set $H= \langle a,b~|~a^{10}=1,b^4=1,bab^{-1}=a^3 \rangle$. $H$ has a normal subgroup $\langle a \rangle$ of order $\le 10$ whose quotient is $\langle b\rangle$ (which has order $\le 4$). Thus any group with these generators and relations as $H$ possess has order at most $40$. Since we already proved that $G$ has order $40$ satisfying these conditions, the abstract presentation of $H$ must equal $G$. It follows that, $G \cong H$.

Have I gotten the solution completely? Any help would be appreciated.


This looks good to me. The canonical map $H \to G$ mapping $a$ to $x$ and $b$ to $y$ is surjective because $G$ is generated by $x$ and $y$ and due to your order argument it has to be bijective, hence is an isomorphism.

Note that generation alone does not suffice in this situation, for example, your group might have more relations than $H$. For instance, the group $$ H' = \langle e,f \:|\: e^{5} = 1, f^4 = 1, fef^{-1} = e^3 \rangle $$ satisfies the relations defining $H$, so it admits a surjective map $H \to H'$ but this one is definitely not an isomorphism.

On the other hand, note that any element of $G$ can be written as $x^iy^j$ in a unique way if we restrict to $0 \leq i \leq 9$ and $0 \leq j \leq 3$. Then we can define a map $G \to H$ by sending $x^iy^{j}$ to $a^ib^j$. This will be a homomorphism and an inverse to the canonical morphism $H \to G$. Uniqueness is important here because otherwise your map might not be well-defined (of course you could also show that $x^{i}y^{j} = x^{i'}y^{j'}$ implies $a^{i}b^{j} = a^{i'}b^{j'}$ for all $i,i',j,j'$ to show that this map is well defined).

  • $\begingroup$ I really like your way that define a map $f: G \to H$ by sending $x^iy^{j}$ to $a^ib^j$ and prove $f$ is an isomorphism. To prove that $f$ is a homomorphism, is it sufficient to prove that $a^{i_1}b^{j_1}a^{i_2}b^{j_2}=a^{i_1+3^{j_1}i_2}b^{j_1+j_2}$? I think I can prove it by using relations of $a$ and $b$. We can also use this equation to prove the surjectivity of $f$, can't we? Finally, $f$ is injective by the observation that $\langle a \rangle \cap \langle b \rangle=\{e\}$. Am I correct? $\endgroup$ – user Sep 23 '15 at 9:38
  • 1
    $\begingroup$ Yes to all but to your last question. More precisely, if you knew that $\langle a \rangle \cap \langle b\rangle $ is trivial then injectivity would follow. However, this is not that clear I believe. Anyway, I would prefer to show bijectivity by noting that $G \to H$ is an inverse of the canonical morphism $H \to G$. $\endgroup$ – Matthias Klupsch Sep 23 '15 at 9:49
  • $\begingroup$ Can you give me the explicit formulation of your canonical map $g:H \to G$? For instance, the map $g$ maps a typical element $a^ib^ja^mb^n...$ (a word) of $H$ to which one of $G$? $\endgroup$ – user Sep 23 '15 at 12:38
  • 1
    $\begingroup$ Your group $H$ is defined by generators and relations, so whenever you have a group, say $K$, and elements $m,n \in K$ such that $m$ and $n$ satisfy the relations of $a$ and $b$, in our case $m^{10} = 1$, $n^4 = 1$ and $nmn^{-1} = m^3$, then there exists a (unique) homomorphism $\varphi: H \to K$, mapping $a$ to $m$ and $b$ to $n$ (this is what I called the canonical map). Because $\varphi$ is a homomorphism, it is clear where a general element $a^{i_1}b^{j_1}a^{i_2}b^{j_2}\dots a^{i_s}b^{j_s}$ is then mapped to (replace $a$ by $m$ and $b$ by $n$). Now apply this to $G=K$, $m=x$, $n=y$. $\endgroup$ – Matthias Klupsch Sep 23 '15 at 13:29
  • 1
    $\begingroup$ Yes, this is correct. The universal property of free groups gives you a morphism of the free group on two generators to $G$ mapping the generators to $x$ and $y$, respectively, and due to the relations satisfied by $x$ and $y$, it factors through $H$ giving you the desired map. $\endgroup$ – Matthias Klupsch Sep 23 '15 at 13:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.