# What familiar group is G isomorphic to?

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.

• 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? – bounceback Dec 4 '19 at 0:16
• 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 $G$? – David G. Stork Dec 4 '19 at 0:17
• @DavidG.Stork, Can you take a look at my answer? – dxdydz Dec 6 '19 at 21:47
• 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. – David G. Stork Dec 6 '19 at 21:49

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.

• Can you take a look at my answer? – dxdydz Dec 6 '19 at 21:46

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.

• Can you take a look at my answer? – dxdydz Dec 6 '19 at 21:46
• @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. – Andreas Blass Dec 6 '19 at 23:39
• isn't that clear and obvious? How should I prove it then? – dxdydz Dec 7 '19 at 3:19
• @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 $a^mb^n=e$. – Andreas Blass Dec 7 '19 at 15:22

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.

• The part where you prove well-definedness is not complete lear. But you can just directly use the defining equations of $G$. – Berci Dec 6 '19 at 21:51
• @Berci, can you be more specific? I don't quite get what part do you think is unclear – dxdydz Dec 7 '19 at 3:17
• How do you know — deep in your heart of hearts — that, e.g., it's not the case that $a^2b^2=e$? – Steven Stadnicki Dec 7 '19 at 5:21
• (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. – Steven Stadnicki Dec 7 '19 at 5:24
• @StevenStadnicki, can you give me any ideas on how to show that $a^mb^n=e \iff m=n=0$? – dxdydz Dec 9 '19 at 5:23

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 $$a^3b^3$$.

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 — 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.