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Let $G$ be the quotient $F_2/\langle a^4,b^4,aba^{-1}b^{-1} \rangle.$
a) What is a simplified form of $ab^8a^5b^{10}$?
b) What is a normal form for the elements of $G$?
c) What familiar group is G isomorphic to?

My attempt:
The quotient is formed by the equivalence relations: $a^4 \equiv e, b^4 \equiv e,ab \equiv ba$
a) $ab^8a^5b^{10}=ab^4b^4aa^4b^4b^4b^2=a^2b^2.$
b) the normal form of elements is $a^ib^j$, where $0 \leq i,j\leq 3$, since if we have degree higher than 3, we can simplify it using the relations $a^4 \equiv e, b^4 \equiv e,ab \equiv ba.$
c) since there are 4 possible choices for $i$ and $j$, I suppose it's isomorphic to $\mathbb Z_4 \times \mathbb Z_4.$ But I don't know how to formally prove that... How to define an mapping $\phi:G\rightarrow \mathbb Z_4 \times \mathbb Z_4$ and prove that it is actually isomorphism

Can somebody check my attempt and help me out with part c)? Thanks in advance.

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    $\begingroup$ Your 3rd relation tells you the group is abelian. The other two tell you the group has 2 generators of order 4... are you familar with the structure theorem? $\endgroup$ – bounceback Dec 4 '19 at 0:16
  • $\begingroup$ a) and b) are correct. Hint: Consider the structure of $\mathbb{Z}_4 \times \mathbb{Z}_4$, such as its generators. Is this the same as $G$? $\endgroup$ – David G. Stork Dec 4 '19 at 0:17
  • $\begingroup$ @DavidG.Stork, Can you take a look at my answer? $\endgroup$ – dxdydz Dec 6 '19 at 21:47
  • $\begingroup$ I didn't check everything carefully, but I don't see any problems with it. One generally approaches by such problems by looking at the structure of the generators. $\endgroup$ – David G. Stork Dec 6 '19 at 21:49
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Your attempt is correct.

Hint for (c): Consider the homomorphism $F_2\to\Bbb Z_4\times\Bbb Z_4$ and prove that its kernel is generated by the given elements.

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  • $\begingroup$ Can you take a look at my answer? $\endgroup$ – dxdydz Dec 6 '19 at 21:46
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Your answer for (b) gives a map $\mathbb Z_4\otimes\mathbb Z_4\to G$, namely $(i,j)\mapsto a^ib^j$, and shows that this map is surjective. The relations in $G$ will make it easy to show that this map is a homomorphism, so what remains is to show it's injective. Injectivity amounts to showing that you normal form in (b) really is a normal form, i.e., that each element of $G$ has a unique representation of that form.

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  • $\begingroup$ Can you take a look at my answer? $\endgroup$ – dxdydz Dec 6 '19 at 21:46
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    $\begingroup$ @dxdydz I think more justification is needed for the step $a^mb^n=1\iff m=0\text{ and }n=0$, specifically for the implication from left to right. $\endgroup$ – Andreas Blass Dec 6 '19 at 23:39
  • $\begingroup$ isn't that clear and obvious? How should I prove it then? $\endgroup$ – dxdydz Dec 7 '19 at 3:19
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    $\begingroup$ @dxdydz If it's clear and obvious, then say why it's obvious. As far as I can see, you need to show (for the left-to-right implication) that, unless $m=n=0$, the relations $a^4\equiv b^4\equiv aba^{-1}b^{-1}\equiv e$ used in defining $G$ don't imply $a^mb^n=e$. $\endgroup$ – Andreas Blass Dec 7 '19 at 15:22
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Thanks everyone for the comments. After reading all the comments to my question, I came up with the following:

Since the normal form of elements of $G$ is $a^ib^j$, where $0≀𝑖,𝑗≀3$, define the mapping $\phi: \mathbb Z_4 \times \mathbb Z_4 \rightarrow G$ by $\phi(([m]_4,[n]_4))=a^mb^n.$ First, if $[m]_4=[c]_4,[n]_4=[d]_4,$ then $\phi(([c]_4,[d]_4))=a^cb^d=a^{[c]_4}b^{[d]_4}$ (since the normal form of elements of $G$ is $a^ib^j$, where $0≀𝑖,𝑗≀3$) $=a^{[m]_4}b^{[n]_4}=\phi(([m]_4,[n]_4))$, so $\phi$ is well-defined.
Next, $\phi(([m]_4,[n]_4)+([c]_4,[d]_4))=\phi(([m+c]_4,[n+d]_4))=a^{m+c}b^{n+d}=a^mb^na^cb^d=\phi(([m]_4,[n]_4))\phi(([c]_4,[d]_4)).$
Therefore, $\phi$ is homomorphism. Now, $\phi(([m]_4,[n]_4))=e_2=1 \iff a^mb^n=1 \iff m=0 \text{ and } n=0.$ Hence, $Ker{\phi}=e_1,$ so $\phi$ is injective.
Finally, for any $a^{m}b^{n} \in G$ there exist $([m]_4,[n]_4) \mathbb Z_4 \times \mathbb Z_4$ such that $\phi(([m]_4[n]_4))=a^{m}b^{n}.$ So, $\phi$ is surjective.
Thus, $\phi$ is isomorphism.

Is this proof valid? Thank you for helping.

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  • $\begingroup$ The part where you prove well-definedness is not complete lear. But you can just directly use the defining equations of $G$. $\endgroup$ – Berci Dec 6 '19 at 21:51
  • $\begingroup$ @Berci, can you be more specific? I don't quite get what part do you think is unclear $\endgroup$ – dxdydz Dec 7 '19 at 3:17
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    $\begingroup$ How do you know β€” deep in your heart of hearts β€” that, e.g., it's not the case that $a^2b^2=e$? $\endgroup$ – Steven Stadnicki Dec 7 '19 at 5:21
  • $\begingroup$ (I should note that the exercise may not be asking you to prove that the group is isomorphic to $C_4\times C_4$, just to claim it and that the first few pieces of the exercise are intended to make that answer probable! While proving it is certainly possible, I think it's somewhat trickier than you may be giving it credit for. $\endgroup$ – Steven Stadnicki Dec 7 '19 at 5:24
  • $\begingroup$ @StevenStadnicki, can you give me any ideas on how to show that $a^mb^n=e \iff m=n=0$? $\endgroup$ – dxdydz Dec 9 '19 at 5:23
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For $c)$, $G$ abelian, order $16$, four elements of order $4\implies G\cong\Bbb Z_4\times\Bbb Z_4$.

To elaborate a little bit, by FTFAG, $G$ is one of $\Bbb Z_{16}, (\Bbb Z_2)^4, \Bbb Z_2\times\Bbb Z_8, \Bbb Z_2\times\Bbb Z_2\times\Bbb Z_4$ or $\Bbb Z_4\times\Bbb Z_4$.

Furthermore, there are clearly at least four elements of order $4$ (actually twelve). Namely $a,b, ab, a^3, b^3, a^2b, ab^2,a^3b,ab^3, a^2b^3, a^3b^2$ and $a^3b^3$.

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As has been noted in various comments, there still need to be some details filled in. Here's how I would handle the piece that you're missing: consider the mapping $\phi$ from an arbitrary word $w$ over $a,b$ and their inverses (in other words, an element of $F_2$) to $C_4\times C_4$ given by counting the number of $a$s and $b$s in $w$ (mod 4, in each case β€” and of course with $a^{-1}$ counting as minus one instance of $a$); you could think of this as the 'charge' of a word. (This is the mapping that Berci mentions in their answer.) Now, show that all of the relations you're given preserve this charge; in other words, if $w$ is a word and $w'$ is another word equivalent to $w$ by one of your group relations, then $w$ and $w'$ have the same charge (i.e., that $\phi(w)$ and $\phi(w')$ are the same element of $C_4\times C_4$). This shows that distinct elements of $C_4\times C_4$ correspond to distinct equivalence classes under your relations, and since you've identified a member of each equivalence class, you have the full isomorphism.

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