If $a,b$ are permutations of $S_8$ and $G=\langle a,b\rangle$ and $N=\langle a^2,b^2\rangle$. How can I prove that $G/N$ is abelian?

Question

If I've got a group $G=\langle a,b\rangle$ where $a$ and $b$ are permutation from $S_8$, and a subgroup of $G$ $N=\langle a^2, b^2\rangle$ (which is in the center of $G$). I have to prove that the quotient group $G/N$ is abelian.

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I know that a group is abelian if the elements of it can be commuted between them. But I don't know how to do it with the quotient subgroup, I mean I don't really know which are the elements from $G/N$ (I think that they are the left cosets but I'm making a mess)

Answer 1

This is false! You cannot expect the result to be true for all subgroups $G$ satisfying your conditions.

Here is a counterexample. Let $a=(1 \ 2)(3 \ 8) (4 \ 7)(5 \ 6), b=(2 \ 8)(3 \ 7)(4 \ 6)$.

Fact. $G=\langle a,b\rangle$ is isomorphic to $D_8$.

To see this, label the vertices of a regular octogon, putting the first one on the $x$-axis of the real plane. If $r$ is the rotation with center $O$ and angle $\dfrac{2\pi}{8}$ and if $s$ is the orthogonal reflection with respect to the $x$-axis, then the permutations induced by $r\circ s$ and $s$ on the eight vertices are exactly $a$ and $b$ (modulo eventual mistakes in my computations...)

Now $D_8=\langle r,s\rangle=\langle rs,s\rangle$. Hence, the group $D_8$, viewed as a subgroup of $S_8$ is $D_8=\langle a,b\rangle$.

To conclude, observe that $a^2=b^2=Id$, so $N=\{Id\}\subset Z(G)$, and $G/N=G/\{Id\}\simeq G$ is not abelian.

So, you have to specify what is the group $G$ you are working with, since the property you are looking for in not true in general.

To answer your original question, as @Ewan suggested, your quotient group (whaterver it is) is generated by the classes $\alpha=aN$ and $\beta=bN$. It is then clear that $G/N$ is abelian if and only if $\alpha\beta=\beta\alpha$. Using the definition fo the group law on $G/N$ (which is well-defined, since a subgroup of the center is normal), you get $abN=baN$, that is $aba^{-1}b^{-1}N=N$, which is equivalent to $aba^{-1}b^{-1}\in N$.

Hence, to prove that $G/N$ is abelian in your specific case, you have to check whether $aba^{-1}b^{-1}\in N$ or not. Unfortunately, since it is false in general and you did not tell who are $G,a$ and $b$, it is not possible to tell you more than that.

Answer 2

Hint: If you put $\alpha=a/N$ and $\beta=b/N$, you have $G/N=<\alpha,\beta>$. So to show that $G/N$ is commutative, it suffices to show that $\alpha\beta=\beta\alpha$. Can you finish from here ?

Answer 3

Let $G$ be a group and $N$ be its normal subgroup. Then the quotient group $\frac{G}{N}=\Big\{gN|g\in G\Big\}$. If group is abelian then quotient group is also abelian. Consider any two element of quotient group and perform.$(aN)(bN)=abN=baN=(bN)(aN)$

Answer 4

I think, you have to identify the group $G=\langle a, b \rangle$ at first, of-course will have several possibility.I am not aware of the general case. I am doing a particular case.

Consider $a=(1)$=identity permutation and $b=(1 \ 2)$ be the two elements of $S_8$. Since $b$ is of order $2$, clearly $G=\langle (1), \ (1 \ 2) \rangle $ forms a group. In that case our $N$ will be $N=\langle a^2, b^2 \rangle =\langle (1) \rangle =\{(1)\}.$ Then, obviously $$G/N=G=\langle (1), \ (1 \ 2) \rangle .$$ Here $G/N$ is cyclic group and hence abelian. Similarly, the same thing happens if $a =(1)$ and $b \in \left\{(1 \ 2)(3 \ 4),\ (1 \ 2) (3 \ 4) (5 \ 6), (1 \ 2) (3 \ 4) (5 \ 6)(7 \ 8) \right\}$ because each $b$ here is of order $2$. Using the same method,you will see that $G/N$ is abelian,in fact cyclic.