Question on finding group isomorphic to $\frac{G}{Z(G)}$

Question

Let $G$ be a group with generators $a$ and $b$ given by $G = \langle a,b:a^4 = b^2 = e, ba = a^{-1}b\rangle$

Let $Z(G)$ denote the centre of the group.

Then I need to show that $\frac{G}{Z(G)}$ is isomorphic to the Klein's four group $K_4$.his

This is a question from my textbook.

I have confusion in this question.

It's no where written that order of $a$ and $b$ are $4$ and $2$ respectively.

If I take both $a$ and $b$ to be identity then $G$ is a trivial group.

If I take just $a$ to be identity, then $G$ is of order $2$.

Then how can the result be true?

Answer 1

Hint: $$G/Z(G)\cong \operatorname{Inn}(G),$$ where $ \operatorname{Inn}(G)$ is the group of inner automorphisms of $G$.

Reference: "Contemporary Abstract Algebra (Eighth Edition)," by Gallian, page 194, Theorem 9.4.

Answer 2

It seems your confusion is less about the computation and more about what "$G = \langle a,b \mid a^4 = b^2 = e, ba = a^{-1}b\rangle$" means, so let me clarify that point for you.

When a group is given by a presentation like $G = \langle a,b \mid a^4 = b^2 = e, ba = a^{-1}b\rangle,$ this means something very specific. Formally, $G$ is the quotient of the free group $F_2$ on two generators $a$ and $b$ by the relations $a^4 = b^2 = e,$ and $abab^{-1} = e.$ That is, let $N$ be the normal closure of the subgroup of $F_2$ generated by $a^4,$ $b^2,$ and $abab^{-1}.$ Then $G$ is defined to be the quotient group $F_2/N.$

In plainer language, writing $G = \langle a,b \mid a^4 = b^2 = e, ba = a^{-1}b\rangle$ means that all elements of $G$ are "words" in $a$ and $b$ (i.e., concatenations $a^{n_1} b^{n_2}\cdots a^{n_{r-1}} b^{n_r},$ where $r\geq 0$ and the $n_i$ are integers), and that you can simplify words as specified by the relations. For example, \begin{align*} a^5b^{-2}a^3b^4ab^{11} &= a^4 a (b^2)^{-1} a^3 (b^2)^2 a (b^{2})^5b\\ &= eae^{-1} a^3 e^2 a e^5 b\\ &= a^4 ab\\ &= ab. \end{align*}

In essence, $G$ is the "free-est" or "largest" group where these relations are satisfied. This is made precise by the following universal property of $G$:

Let $G = \langle a,b \mid a^4 = b^2 = e, ba = a^{-1}b\rangle.$ Let $H$ be a group such that there exist $\alpha,\beta\in H$ such that $\alpha^4 = \beta^2 = e,$ $\beta\alpha = \alpha^{-1}\beta,$ and that any $h\in H$ can be written as a word in $\alpha$ and $\beta.$ Then there exists a unique group homomorphism $f : G\to H$ such that $f(a) = \alpha$ and $f(b) = \beta.$ Moreover, $f$ is a surjection.

This follows from the definition of $G$ as a quotient of the free group on $a$ and $b$ and the universal property of quotient groups. The fact that $f$ is a surjection follows from the fact that any element of $H$ can be written as a word in $\alpha$ and $\beta,$ and that both $\alpha$ and $\beta$ are in the image of $f.$

Here's the key point: the presentation given uniquely specifies the group $G$ -- you are not working with an arbitrary group $G$ such that those relations are true. So, you may not simply assume that the order of $a$ is $1,2,$ or $4,$ and the the order of $b$ is $1$ or $2.$ The order of $a$ and $b$ are what they are, and are determined uniquely by the presentation.

To do the problem, you can find a description of $G$ and explicitly compute $Z(G)$ and find the quotient. The group $G$ is rather small, so the computation is quite manageable.

If you're not sure how to proceed with this computation, here's a starting point: notice that the relation $ba = a^{-1}b$ lets you move $a$'s and $b$'s past each other in a word, at the cost of an $a^{-1}.$ This lets you write any word in the form $a^n b^m,$ where $n,m\in\Bbb{Z}.$ Now, try to find a set $S\subseteq\Bbb{Z}^2$ such that every element of $G$ can be written uniquely as $a^n b^m$ with $(n,m)\in S.$

Let me just end with a word about the word problem. The word problem is loosely the following: given a presentation of a group as a set of generators and relations $G = \langle S\mid R\rangle,$ how can we determine if $G$ is the trivial group? This is a very difficult problem (in fact, unsolvable in some generality), and has spawned a lot of research in group theory!

Answer 3

It is possible to "reverse engineer" both the solution and the group $G$ given JCAA's comment (that $o(a)$ divides $4$ and $o(b)$ divides $2$) and the assumption that the question is correct. This allows us to bypass the rather subtle theory of presentations.

1. The relation $ba=a^{-1}b$ means that every element of the group can be written in the form $a^nb^m$. As $o(a)$ divides $4$ and $o(b)$ divides $2$, this means that $|G|$ divides $8$.
2. As $|G/Z(G)|=4$, we have that $4$ divides $|G|$. Hence, $|G|\in\{4, 8\}$.
3. If $|G|=4$ then $G$ is abelian, and so $G/Z(G)$ is trivial. Hence, $|G|=8$.

We can now reconstruct $G$: The fact that $|G|=8$ means that $\langle a\rangle\cap\langle b\rangle=1$, while $\langle a\rangle$ is normal in $G$ (it has index two, or alternatively apply the relator $ba=a^{-1}b$). Hence, $G\cong\mathbb{Z_4\rtimes Z_2}$.

We can also see how the exercise was meant to go:

1. As with (1) above, $|G|$ divides $8$.
2. Spot that $\mathbb{Z_4\rtimes Z_2}$ satisfies the relations, and so in fact $G\cong\mathbb{Z_4\rtimes Z_2}$.
3. As $G$ is non-abelian, $G/Z(G)$ is non-cyclic (this is a standard exercise).
4. Hence, either $Z(G)$ is trivial or $G/Z(G)$ is the Klein $4$-group.
5. Conclude that $Z(G)$ is non-trivial (which I shall leave to you :-) ), and so $G/Z(G)$ is the Klein $4$-group.