I was given the following two problems. Is my solution correct?
(1) Let G = $\langle a,b | a^7 = b^3 = 1, bab^{-1} = a^2\rangle$
Show this is isomorphic to $H = C_7 \rtimes C_3$ for some $\phi: C_3 \rightarrow Aut(C_7)$. Give explicitly what $\phi$ is.
Here $C_i$ represents the cyclic group of order i.
By the universal property we have a homomorphism $f: G \rightarrow H$ provided that we can map the generators in G to elements in H such that the relations are satisfied.
Clearly, we map $a \rightarrow (1,0)$ and $b \rightarrow (0,1)$ which satisfies the relations $a^7 = b^3 = 1$ since the identity is $(0,0)$ for H.
Now, the last relation allows us to determine $\phi$ explicitly as follows:
To calculate $a^2$ in H we have:
$(1,0)(1,0) = (1+1,0) = (2,0)$
To calculate $bab^{-1}$ we have:
$(0,1)(1,0)(0,1)^{-1} = (0 + \phi(1)(1), 1+0)(0, 2) = (\phi(1)(1), 1)(0, 2) = (\phi(1)(1) + \phi(1)(0), 1+2) = (\phi(1)(1), 0)$
where we have used the fact that $\phi(1)(0) = 0$ and $(0,1)^{-1} = (0,2)$.
Thus, we must have $\phi(1)(1) = 2$. But this completely determines $\phi$ since $C_3$ and $C_7$ are both cyclic. So we have a homomorphism f between G and H, as desired.
Now, to show it's an isomorphism, we would need to show f is injective since it's clearly onto (since f maps to the generators). But how do I show this part?
(2) For G = $\langle a,b | a^7 = b^3 = 1, bab^{-1} = a^3\rangle$, determine what group it is isomorphic to.
The hint given was to simplify the relations.
I did this as follows:
$bab^{-1} = a^3 \implies ba^3b^{-1} = a^2 \implies b^2ab^{-2} = a^2 \implies b^2a^3b^{-2} = a^6 \implies b^3ab^{-3} = a^6 \implies a = a^6 \implies a^2 = 1$ we cubed both sides, subbed in, cubed again, etc.
But $a^7 = a^2 = 1 \implies a = 1$. Thus, the presentation reduces to one generator b with the relation $b^3 = 1$. Thus it must be the case that $G \simeq C_3$.
Could it also be the case that G is the trivial group as well? Or do we just exclude this possibility?
Best Answer
Perhaps all of what I am going to say is already obvious to you, still.
$\newcommand{\Set}[1]{\left\{ #1 \right\}}$Consider the first group. You have proved that there is a surjective homomorphism $f : G \to H$. So you know that $G$ has order at least $21$ (at this stage, it could conceivably be infinite).
Now if we prove that $G$ has order at most $21$, we're done, because $f$ will turn out to be a surjective map between two finite sets with the same number of elements, and thus $f$ is bijective.
To show this, consider the subset $$ S = \Set{ a^{j} b^{i} : 0 \le i < 3, 0 \le j < 7 } $$ of $G$. This has order at most $21$. We will show that this is a subgroup of $G$. Since it contains the generators $a, b$ of $G$, we have $S = G$, and you're done.
Let $a^{j} b^{i} , a^{n} b^{m} \in S$, and consider the product $$ a^{j} b^{i} \cdot a^{n} b^{m} = a^{j + 2^{i} n} b^{i+m}, $$ where we have repeatedly used the fact that $$ b a^{x} = b a^{x} b^{-1} b = (b a b^{-1})^{x} b = a^{2 x} b. $$ Using the relations $a^{7} = 1$ and $b^{3} = 1$, we see that the product lies in $S$. A similar argument applies to the inverse; or you can reduce the calculation of the inverse to that of the product: consider $a^{j} b^{i} \in S$, then $$ (a^{j} b^{i})^{-1} = b^{-i} a^{-j} = b^{m} a^{n} = (a^{0} b^{m}) \cdot (a^{n} b^{0}) \in S, $$ where $0 \le m < 3, 0 \le n < 7$ are obtained from the relations $a^{7} = 1$ and $b^{3} = 1$.