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Let G be the quaternion group and Z(G) is center of the group G then the quotient group G/Z(G) is isomorphic to-

A) G

B) ({1, -1}, .)

C) Klein's 4-group

D) Group of integers modulo 4

  • I've figured out that option A and B are wrong. Given quotient group is abelian group of order 4.

-Group in Option A) isn't abelian.

-Group in option B) is of order 2.

Now to choose right one from C and D, i need to know that given quotient group is cyclic or not? But How to do that?

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    $\begingroup$ If you have the theorem that the only time the quotient by the center is cyclic is when it’s trivial, then you can also discard d. $\endgroup$ – Randall Jul 1 '19 at 19:46
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You can compute the center and you will get that it is a group of order $2$ such that the quotient has order $4$ as you mentioned. Therefore C) or D) could be correct by now. If you are considering the elements in the quotient you will see that every non-trivial element has order $2$ and therefore you will get the Klein-$4$-group. So yes, check whether the quotient is cyclic by computing the orders.

Let me give an example. The order of $i \cdot \lbrace 1,-1 \rbrace \in Q_8/Z(Q_8)$ is $2$, as $i \not\in \lbrace 1,-1 \rbrace$ and

$(i \cdot \lbrace 1,-1 \rbrace)^2 = i^2 \cdot \lbrace 1,-1 \rbrace = -1 \cdot\lbrace 1,-1 \rbrace = 1 \cdot \lbrace 1,-1 \rbrace$.

Now do the computation for the other $2$ non-trivial elements of the quotient.

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    $\begingroup$ Thank you so much. I absolutely got it. After computing the order of other 2 non trivial elements, it's clear that option C is correct. $\endgroup$ – user684646 Jul 1 '19 at 19:55
  • $\begingroup$ Yes, exactly. Glad that I could help. $\endgroup$ – TMO Jul 1 '19 at 19:56
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One way to do this is to consider the presentation of the group. Note that

$$Q_8 = \langle i,j,k~|~i^2=j^2=k^2=ijk=-1\rangle.$$

Since $Z(W_8) = \{1,-1\}$, in $Q_8/Z(Q_8)$ we have that $ijk = 1$, or $ij = k$. Thus, we know that $(ij)^2 = 1$ in the quotient, implying that $ij=ji$. Thus, we have a presentation

$$Q_8/Z(Q_8) = \langle i,j~|~i^2 = j^2 = 1,~ij=ji\rangle.$$

This is exactly the presentation of the Klein four group.

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  • $\begingroup$ Thanks a lot to show that structure is same as of klein 4 group. I got the answer. $\endgroup$ – user684646 Jul 1 '19 at 19:58

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