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I am working through Rotman's abstract algebra and the above question has been giving me problems. I have a solution book for it which gives the following:

We know $Z(Q) = \{\pm E\}$ so $Q(Z)$ has order $4$. We also know that for all $M \neq \pm I$, $M^2 = I$. Thus every nonidentity element of $Q/Z(Q)$ has order $2$ and hence $Q/Z(Q) \cong V$.

My question is how does the last step follow? Every element having order $2$ means it's abelian but even if all elements have the same order in $V$ and $Q/Z(Q)$ it does not mean that they are isomorphic. So how is proving the order sufficient?

Shaun
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P4PL4
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    Since you know it's abelian, it's either $\mathbb{Z}/4\mathbb{Z}$ or $(\mathbb{Z}/2\mathbb{Z})^2$ (by the fundamental theorem of finite abelian groups). Only the latter has no elements of order $>2$. – Noah Schweber Jul 27 '22 at 20:22
  • @NoahSchweber Thank you for your swift reply! I have not yet covered the fundamental theorem of abelian groups. So I am assuming both V and Q/Z(Q) are isomorphic to that latter group and thus are isomorphic since isomorphism forms an equivalence class on groups? Thanks again for your help – P4PL4 Jul 27 '22 at 20:26
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    You know $Q/Z(Q)$ has four elements, is abelian, each nonidentity element is of order $2$. You also know the multiplication table, since it is inherited from $Q$. It should be pretty obvious that this is isomorphic to $V$. What Noah Schweber is noting is that there are in fact just two groups of order $4$ up to isomorphism: the cyclic one and $V$, and this is clearly not cyclic. – Arturo Magidin Jul 27 '22 at 20:30
  • @ArturoMagidin Ah I understand, so the solution merely omitted the step of providing a multiplication table as trivial. I thought that this was a full proof provided and was confused if that was a result that held in general. Thank you for also clarifying Noah Schweber's answer! – P4PL4 Jul 27 '22 at 20:36
  • @P4PL4 Another way to see it is as follows - $Q/Z(Q)$ is a group of order $4$, meaning either $\mathbb{Z}/4\mathbb{Z}$ or $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$. Prove the following theorem (not hard) - if $G/Z(G)$ is cyclic, then $G$ is abelian. Now, since $Q$ is not abelian, we have that $Q/Z(Q)$ cannot be cyclic, meaning $Q/Z(Q)\cong \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ – Math101 Jul 27 '22 at 20:53
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    I couldn't help but refine "the quaternions" (which might often mean the Hamiltonian quaternions...) to "the quaternion 8-group"... which is apparently the intention. – paul garrett Jul 27 '22 at 21:06
  • Instead of the fundamental theorem of finite abelian groups we could also use linear algebra to show abelian groups of prime exponent are elementary abelian. But showing there are only two groups of order four in particular is trivial in comparison. – anon Jul 30 '22 at 09:46

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By Lagrange's Theorem,

$$\begin{align} |Q/Z(Q)|&=|Q|/|Z(Q)|\\ &=8/2\\ &=4. \end{align}$$

As noted in the comments, there is only two groups of order four up to isomorphism: the Klein four group $V$ and $\Bbb Z_4$. But the latter has an element of order four; therefore, $Q/Z(Q)\cong V.$

Shaun
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