What familiar group is G isomorphic to?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $a^3b^3$.

Answer 4

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 $0≤,≤3$, 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 $0≤,≤3$) $=a^{[m]_4}b^{[n]_4}=\phi(([m]_4,[n]_4))$, 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.

Answer 5

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.