# Thread: Quotient groups and centers

1. ## Quotient groups and centers

i) Prove that $Q/Z(Q) \cong V$, where $Q$ is the group of quaternions and $V$ is the four-group. Conclude that the quotient of a nonabelian group by its center can be abelian.

ii) Prove that $Q$ has no subgroup isomorphic to $V$. Conclude that the quotient $Q/Z(Q)$ is not isomorphic to a subgroup of $Q$.

For part two, is it enough to say

An isomorphism preserves the order of all the elements. Therefore $Q$ has only one element of order 2, which means every subgroup of $Q$ will have a maximum of 1 element of order 2, which is $A^2$ also known as $-I,$ because $V$ consist of 1 element of order 1,(the identity) and 3 elements of order 2, it is impossible to make a subgroup of $Q$ with 3 elements of order 2. Therefore $Q$ has no subgroup isomorphic to $V$.

If that is true, then it follows that $Q/Z(Q)$ is not isomorphic to a subgroup of $Q$ because, $Q/Z(Q) \cong V$, and we know that no subgroup of $Q$ is isomorphic to $V$, then it follows $Q/Z(G)$ is not isomorphic to a subgroup of $Q$.

Side Note: I understand if this is the worst mathematical proof one has ever witness.

2. $Q$ is a quaternion group, not group of quaternions. The quaternions form a division ring whose additive group is an infinite Abelian group – whereas $Q$ is a finite nonabelian group of order $8.$ The two are different.

(i)

Use the result that if $Q/Z(Q)$ were cyclic, then $Q$ would be Abelian (contradiction).

(ii)

$Q$ is generated by two elements $x,y$ with $x^2=y^2$ and $y^{-1}xy=x^{-1}.$ The only element of order $2$ is $x^2\,(=y^2)$ and so $Q$ cannot have a Klein subgroup (which would require three elements of order $2).$

3. In my book the definition is: The group of quaternions is the group Q of order 8 consiting of the matrices in $GL(2,\bold C)$

so are u using the same group... $Q$ is, a finite group, nonabelian of order 8... namely

$Q=(I, A,A^2,A^3,B,BA,BA^2,BA^3)$

4. Originally Posted by ux0
In my book the definition is: The group of quaternions is the group Q of order 8 consiting of the matrices in $GL(2,\bold C)$
Very well. But I still think the term quaternion group is much preferable.

Originally Posted by ux0
$Q$ is, a finite group, nonabelian of order 8... namely

$Q=(I, A,A^2,A^3,B,BA,BA^2,BA^3)$
That’s not the complete description of the group – you still have to add that $B^2=A^2.$ This is to distinguish it from the other nonabelian group of order $8,$ namely the dihedral group, in which $B^2=I.$ The fact that $B^2=A^2$ means that $A^2$ is the only element in the group of order $2.$

By the way, are you familiar with this result? If $G$ is a group such that $G/Z(G)$ is cyclic, then $G$ is Abelian. This is used to prove part (i). (Also note that $Z(Q)=\{I,\,A^2\}.)$

5. Originally Posted by proscientia

That’s not the complete description of the group – you still have to add that $B^2=A^2.$ This is to distinguish it from the other nonabelian group of order $8,$ namely the dihedral group, in which $B^2=I.$ The fact that $B^2=A^2$ means that $A^2$ is the only element in the group of order $2.$
If you take any term in $Q \neq \{I,\,A^2\}$, and Square it, you will get $-I$, therefrom every term but these two, are of order 4..

Originally Posted by proscientia
By the way, are you familiar with this result? If $G$ is a group such that $G/Z(G)$ is cyclic, then $G$ is Abelian. This is used to prove part (i). (Also note that $Z(Q)=\{I,\,A^2\}.)$

$\forall a,b\in G\;\exists n_a,k_b\in \mathbb{Z}\,\,s.t.\,\,a=x^{n_a}z_a\,\,b=x^{k_b}z_b \,\,\,with\,\;z_a,z_b\in Z(G)$

$ab= x^{n_a}z_ax^{k_b}z_b
$

$=x^{n_a}x^{k_b}z_az_b
$

$=x^{n_a+k_b}z_az_b
$

and

$ba=x^{k_b}z_bx^{n_a}z_a
$

$=x^{k_b}x^{n_a}$ $z_bz_a
$

$=x^{k_b+n_a}z_bz_a
$

$=x^{n_a+k_b}z_az_b$

ab=ba

Therefore it is abelian

Not really sure how that works for part one tho...

$A,B \in Q$

$A=\begin{bmatrix}
0 & 1 \\ -1
& 0
\end{bmatrix}$

$

B=\begin{bmatrix}
0 & i \\ i
& 0
\end{bmatrix}$

$Z(Q)=\{I,\,A^2\}.$

6. Originally Posted by ux0
i) Prove that $Q/Z(Q) \cong V$, where $Q$ is the group of quaternions and $V$ is the four-group. Conclude that the quotient of a nonabelian group by its center can be abelian.

ii) Prove that $Q$ has no subgroup isomorphic to $V$. Conclude that the quotient $Q/Z(Q)$ is not isomorphic to a subgroup of $Q$.

For part two, is it enough to say

An isomorphism preserves the order of all the elements. Therefore $Q$ has only one element of order 2, which means every subgroup of $Q$ will have a maximum of 1 element of order 2, which is $A^2$ also known as $-I,$ because $V$ consist of 1 element of order 1,(the identity) and 3 elements of order 2, it is impossible to make a subgroup of $Q$ with 3 elements of order 2. Therefore $Q$ has no subgroup isomorphic to $V$.

If that is true, then it follows that $Q/Z(Q)$ is not isomorphic to a subgroup of $Q$ because, $Q/Z(Q) \cong V$, and we know that no subgroup of $Q$ is isomorphic to $V$, then it follows $Q/Z(G)$ is not isomorphic to a subgroup of $Q$.

Side Note: I understand if this is the worst mathematical proof one has ever witness.
Part two is fine. For your part one, note that the center has order 2, and so the quotient has order 4. There are only 2 groups of order 4, so it is enough to find two distinct non-trivial elements in the quotient group that are of order 2.

7. Originally Posted by Swlabr
For your part one, note that the center has order 2, and so the quotient has order 4. There are only 2 groups of order 4, so it is enough to find two distinct non-trivial elements in the quotient group that are of order 2.

So, we know there is only 2 groups of order 4.. $V$ and $ =1$ (I'm trying to say the cyclic group of order 4)...

But The quotient group I'm getting is

$I = \begin{bmatrix}
1 &0 \\
0& 1
\end{bmatrix}$

$
A= \begin{bmatrix}
0 &-1 \\
1& 0
\end{bmatrix}$

$B=\begin{bmatrix}
-1 &0 \\
0& -1
\end{bmatrix}$

$C= \begin{bmatrix}
0 &1 \\
-1& 0
\end{bmatrix}$

Which is orders 1,4,2,4 respectively... Is it so non-trivial i am looking right pass it?? But to be isomorphic i need my quotient group elemtns to be order 1,2,2,2, seeing that is the order of V elements..

Is it possible to say my quotient group .. $Q/Z(Q) =$

$I
$

$\begin{bmatrix}
1&0 \\
0& -1
\end{bmatrix}$

$\begin{bmatrix}
-1&0 \\
0& 1
\end{bmatrix}$

$\begin{bmatrix}
-1&0 \\
0& -1
\end{bmatrix}$

This would get me, orders 1,2,2,2.. which would make it isomorphic to V

8. Neither of these groups are your quotient groups -

$\begin{bmatrix}
-1&0 \\
0& -1
\end{bmatrix} \equiv I \text{ mod } Z(G)$
...

You need to find 4 cosets each of two elements such that $gN=hN \Leftrightarrow gh^{-1} \in N$ where $N=Z(G)$...one of these will be the center, the other ones will be your non-central matrices paired with their negatives (although that is just off the top of my head - I'll leave it to you to check properly!).

So, for instance, $\{AZ(G), -AZ(G)\}$ would be one coset.

9. Why is everybody still ignoring what I wrote?

Originally Posted by proscientia
If $G$ is a group such that $G/Z(G)$ is cyclic, then $G$ is Abelian. This is used to prove part (i). (Also note that $Z(Q)=\{I,\,A^2\}.)$
$Q/Z(Q)$ is a group of order $4.$ If $Q/Z(Q)\not\cong V,$ then $Q/Z(Q)$ would be cyclic, which would mean $Q$ was Abelian (contradiction).

10. Originally Posted by proscientia
Why is everybody still ignoring what I wrote?

$Q/Z(Q)$ is a group of order $4.$ If $Q/Z(Q)\not\cong V,$ then $Q/Z(Q)$ would be cyclic, which would mean $Q$ was Abelian (contradiction).
I have no idea! I don't know how I missed it - sorry!

It's a nice way though.

11. Wow, i finally get it.. because it is a group of order 4, and its not a cyclic one, then it has to be V or a isomorphism of V.. I literary spent hours trying to figure out why that contradiction proved that.