discussion of quaternion team of order 8

Let G be the team specified by generators a, b and also relationships $a^4=e$, $a^2b^{-2}=e$, $abab^{-1}=e$. Given that the quaternion team of order 8 is created by components $a,b$ pleasing the previous relationships, there is an epimorphism from $G$ onto $Q_8$. Allow $F$ be the free team on $\{a,b\}$ and also $N$ the regular subgroup created by $\{ a^4, a^2b^{-2}, abab^{-1} \}$. Just how to list the regular subgroup N and also share the team F/N?

Many thanks

2019-05-18 22:22:59
Source Share
Answers: 2

Two factors concerning this. As a subgroup of a free team, N is itself a free team. There is a mathematical procedure for locating a free set of generators of N, called the Schreier generators. An instance of this existed carefully in a current thing "Normal Closure in teams II":

Normal closure in groups II

Using a computer system program (MAGMA), I obtained the adhering to set of 9 generators of N:

N.1 = b * a * b * a^-1
N.2 = b^2 * a^-2
N.3 = a^-1 * b^2 * a^-1
N.4 = b^-1 * a * b^-1 * a^-1
N.5 = a^4
N.6 = a^2 * b^2
N.7 = a * b * a * b^-1
N.8 = a * b^2 * a
N.9 = a * b^-1 * a * b

The 2nd factor is that the first relationship $a^4=e$ in your discussion is in fact repetitive. The quaternion team is specified by the discussion with 2 generators $a,b$ and also simply both relationships $a^2=b^2$ and also $abab^{-1} = e$. It is a wonderful workout to confirm that!

2019-05-21 21:25:18

To list the team $F/N$, the most effective method this instance is to locate a regular kind. Keep in mind that the last relationship $abab^{-1}=e$ suggests that $ba=a^{-1}b = a^3b$ (that is, the component $b^{-1}a^{-3}ba$ remains in $N$) ; so any kind of coset rep which contains the string $ba$ can be changed with a coset rep in which the string $ba$ is changed with the string $a^3b$. The reality that $a^2b^{-2}\in N$ suggests that any kind of coset rep that in which you have a $b^2$ can be changed with one that has an $a^2$ rather. Continuing this way, you can end that every coset can be stood for by a component of the kind $a^i b^j$ with $0\leq i\leq 3$ and also $0\leq j\leq 1$ (you can get rid of any kind of $a^4$ given that $a^4\in N$). Hence, the team $F/N$ contends the majority of $8$ components. You currently recognize it contends the very least $8$ components, to make sure that suggests that each of these components stand for distinctive cosets of $N$, to make sure that offers you a means to share $F/N$ making use of the reps.

" Writing down" the team $N$ is a little bit harder, given that it is a boundless subgroup in $F$. Just what do you suggest by "listing $N$"? A means to establish if an offered component of $F$ hinges on $N$? This can be attained by locating its "coset rep" from amongst the unique set recognized over. If you get $a^0b^0$, after that the component remained in $N$ ; otherwise, after that it was not.

2019-05-21 06:04:25