Let Q8 represent the group of quaternions and V4∼= Z/2Z ⊕ Z/2Z represent the Klein four-group.

Question

Show that Q8/Z(Q8) ~= V4 where Z(Q8) is the center of Q8.

Conclude that the quotient of a non-Abelien group by its center can be Abelian.

I have very little experience with Q8 and V4 and I was wondering if someone could clarify this problem for me.

Answer 1

The first step is to find $Z(Q_8)$. $Q_8$ has $8$ elements, $V_4$ has $4$ elements. So seeing the problem, by the fact that if $N \triangleleft G$, $|G/N|$= $\frac{|G|}{|N|}$, you suspect $Z(Q_8)$ to have $2$ elements. Of course, $1 \in Z(Q_8)$. By direct computation one finds $-1 \in Z(Q_8)$. Because $i, j$ and $k$ do not commute with each other, we have $i, j, k \notin Z(Q_8)$. Now it follows $-i, -j, -k \notin Z(Q_8)$, because otherwise we could multiply by $-1 \in Z(Q_8)$ and using the fact that $Z(Q_8)$ is a subgroup we would have $i, j, k \in Z(Q_8)$.

Now $G:=Q_8/Z(Q_8)$ has $4$ elements, and there are, up to isomorphism, only $2$ groups of order $4$, one of which is $\mathbb{Z}_{4}$ and the other is $V_4$. $G$ has $3$ elements of order $2$, namely $\bar{i}, \bar{j}$ and $\bar{k}$, so $G \cong V_4$. Intuitively (and one can prove it based on this intuition), when you quotient out $\{1, -1 \}$, you force $-1$ to being equivalent to $1$, so that firstly all the noncommutativity vanishes, because for example $ij=-ji$, and secondly $i^2=j^2=k^2=-1$ becomes equivalent to $1$.

Your little experience of $V_4$ should not be a problem, because all you need to know about $V_4$ is the fact it has $4$ elements, $1$ of which of course is the identity and $3$ of which have order $2$, and two different nonidentity elements of $V_4$ multiplied by eachother gives you the third nonidentity element.

Answer 2

Here is a nice argument using the following general principles:

1. Recall that a $p$-group is a group whose order is a power of the prime $p$. Any nontrivial $p$-group has a nontrivial center.

2. Let $G$ be a group such that the quotient $G/Z(G)$ of $G$ by its center is cyclic. Then $G$ is commutative.

Note that $Q_8$ is a $2$-group; its order is equal to $8=2^3$. By Lagrange, the order of the center $Z(Q_8)$ divides $8$, i.e., $\# Z(Q_8)=1,2,4$ or $8$. By statement 1, the center $Z(Q_8)$ is nontrivial. Hence $\# Z(Q_8)\neq1$. Since $Q_8$ is not commutative, one has $\#Z(Q_8)\neq8$, but also $\#Z(Q_8)\neq 4$ by statement 2 as any group of order $2$ is cyclic. Therefore, $\#Z(Q_8)=2$. Again by statement 2, the quotient $Q_8/Z(Q_8)$ cannot be cyclic. Since this quotient has $4$ elements it has to be isomorphic to the Klein group.

For completeness, here are proofs of statements 1 and 2 above:

Proof of 1: Let $G$ be a nontrivial $p$-group and let $G$ act on itself by conjugation. Since a nonsingleton orbit has cardinality divisible by $p$, and since the set $G$ on which $G$ acts has cardinality divisible by $p$ as well, the number of singleton orbits is divisible by $p$ too. Since the singleton orbits are exactly the elements of the center of $G$, one has that $\#Z(G)$ is divisible by $p$. It follows that the group $Z(G)$ is nontrivial.

Proof of 2: let $g$ in $G$ be a generator of the quotient $G/Z(G)$. Any element of $G$ can be written in the form $g^nz$ with $n\in\mathbf Z$ and $z\in Z(G)$. Two such elements $g^nz$ and $g^mw$ commute since $$ g^nzg^mw=g^ng^mzw=g^{n+m}wz=g^{m+n}wz=g^mg^nwz=g^mwg^nz, $$ since $w$ and $z$ are in the center of $G$.