Suppose that group $G$ has a presentation $\langle a,b\mid a^{4} = b^{7} = 1, ab=ba^{3}\rangle$

Question

Suppose that group $G$ has a presentation $\langle a,b \mid a^{4}= b^{7}=1, ab=ba^{3}\rangle$. Show that the order of $G$ is $28$.

I can only try combining and permuting $a$ and $b$ and eliminating them by the presentation of $G$ to get the answer. It's tough work. I think there must be some easy and 'systematic' metheds to solve the question. Any ideas?

Answer 1

Let us show that the order of $G$ is $\color{red}{14}$, and that it is commutative.

From $ab=ba^3$ we get $aba=ba^3\ a=ba^4=b$. This then gives $a^2ba^2= a(aba)a=aba=b$. Then $$ \begin{aligned} 1 &= b^7\\ &=(aba)^7\\ &=aba\ aba\ aba\ aba\ aba\ aba\ aba\\ &=ab\ a^2ba^2\ b \ a^2ba^2\ b\ a^2ba^2\ ba\\ &=ab\ b\ b\ b\ b\ b\ ba\\ &=ab^7a\\ &=aa\ . \end{aligned} $$ (I hope i typed it correctly, each line containing the many $b$'s should have seven of them in the corresponding non-commutative product.) So we in fact have $a^2=1$.

Now we are following user1729...

So $ab=ba^3=ba\ a^2=ba$, and $G$ is commutative. There is only one such structure, $G\cong\Bbb Z/2\times\Bbb Z/7\cong\Bbb Z/14$.

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Computer check added... Well, i was also curious to see the opinion of sage on the above...

sage: F.<a,b> = FreeGroup()
sage: G = F / [a^4, b^7, a*b*a*b^-1]
sage: G.is_finite()
True
sage: G.order()
14
sage: G.is_commutative()
True

Next time i will try this first...