# Quaternions Group Homomorphic to Klein Group

I'm currently studying some stuff about group theory and I came to problem of showing that $$\displaystyle\frac{Q_8}{\langle-1\rangle}\cong V_4,$$ so I checked on this link: Quaternions Group and Klein Group, which seems to clarify somehow what I wanted to know. But now I'm curious about how to prove the statement using the first isomorphism theorem. Here is what I have: Setting the map $\varphi:Q_8\to V_4$ such that $\varphi(a)=|a|$ we find that $\varphi$ is a morphism of groups with $Ker(\varphi)=\langle -1\rangle$, so applying the theorem the statement holds. My question is about the legality of $\varphi$, I think it's ok but I'd like another opinion, approach, thought about it.

• You could do it less directly as well. Note that $Q_8/\langle -1\rangle$ has four elements and every non-identity element has order $2$. – Arthur Aug 5 '16 at 18:18
• What do you mean by $|a|$ in this context? – carmichael561 Aug 5 '16 at 18:18
• @carmichael561 I mean the absolute value. This was my very first concern about it, because I was curious about how to define the morphism properly – Cristian Baeza Aug 5 '16 at 19:29
• Notice that $|a| = 1$ for all $a$, so your morphism is trivial (furthermore, it is not surjective either). – Alex M. Aug 5 '16 at 19:52
• Yeah, no, well, there it is, I didn't mean the norm of $a$, I meant the "positive" version of $a$. I wasn't sure how to exprese it, and now I'm wondering how. – Cristian Baeza Aug 5 '16 at 19:59

## 3 Answers

Well, if $Q_8 = \{ \pm 1, \pm i, \pm j, \pm k \}$ with $(-1)^2 = 1$ and $i^2 = j^2 = k^2 = -1$, and the Klein group is $V_4 = \{ 1, x, y, xy \}$, then define $\varphi (i) = x, \ \varphi (j) = y$ and obtain $\varphi(1) = 1$ (by definition) and $\varphi (k) = \varphi (ij) = \varphi (i) \varphi (j) = xy$. It also follows that $\varphi (-1) = \varphi (i^2) = \varphi (i) ^2 = x^2 = 1$, hence $\varphi (-i) = \varphi (-1) \varphi (i) = \varphi (i)$ etc.

It follows that $\ker \varphi = \{-1, 1\}$, so that $\frac {Q_8} {\langle -1 \rangle} \simeq V_4$.

• That's even better than what I was trying to reach. Thanks. – Cristian Baeza Aug 5 '16 at 19:26

You can prove that $\displaystyle\frac{Q_8}{V_4}\cong\langle-1\rangle$ . Now you know that $V_4$ and $Q_8$ is of order 4 and 8 respectively . So their quotient group is of order 2 and there is only one group (up to isomorphism ) of order i.e $\langle-1\rangle$ . so the quotient group must be isomorphic to $\langle-1\rangle$ .

Edit : This proof doesn't make much sense . It is not always true that $G/H \simeq K$ , then $G/K \simeq H$ . Most of the time $G/K$ is not group. Take for example $G=S_3$ ,$H=\langle\alpha\rangle$ and $K=\langle\beta\rangle$ where $\alpha^3=\beta^2=1$, $\alpha,\beta$ elements in $G$.

• This can't be possible. In general, if $G/H \simeq K$ and $G/K \simeq H$, then $G \simeq H \times K$. We already know that $Q_8 / \langle -1 \rangle \simeq V_4$. If it were true that $Q_8 / V_4 \simeq \langle -1 \rangle$, as you claim, then $Q_8 \simeq V_4 \times \langle -1 \rangle \simeq \Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_2$. One consequence of this is that all elements are of order $2$, and that $Q_8$ is commutative - clearly not true. Therefore, I believe your claim to be false. – Alex M. Aug 5 '16 at 19:46
• Yes you are right . I have corrected the mistakes now. – Suman Kundu Aug 5 '16 at 21:29

There are, up to isomorphisms, two groups of order $4$: the cyclic group and the Klein group $V_4$. For any $x\in Q$, $x^2\in\{1,-1\}$, so every element in $Q/\{1,-1\}$ has order dividing $2$. Hence $Q/\{1,-1\}$ cannot by cyclic.

• Absolutely true, but the OP wanted a proof based on the first isomorphism theorem for groups. – Alex M. Aug 6 '16 at 20:55