Which group is this quotient group isomorphic to?












0












$begingroup$



Let $$G = left{begin{pmatrix} q&0\a+bi&qend{pmatrix} mid q in mathbb{Q}^ast, a,binmathbb{R}right}$$
and $$H = left{begin{pmatrix} q&0\a+ai&qend{pmatrix} mid q in mathbb{Q},q>0, ainmathbb{R}right}$$
Which known group is the quotient group $(G, ·)/H$ isomorphic to?




I don't know how to solve this. What I've tried:



$A,B in G$ belong to the same coset iff $A^{-1}B in H$.
Let $A = begin{pmatrix} q&0\a+bi&qend{pmatrix}$ and $B = begin{pmatrix} r&0\c+di&rend{pmatrix}$. Then $A^{-1} = begin{pmatrix} 1/q&0\(-a-bi)/q^2&1/qend{pmatrix}$ and $A^{-1}B = begin{pmatrix} r/q&0\frac{qc-ra+i(qd-rb)}{q^2}&r/qend{pmatrix}$



That means that $qc-ra$ should be equal to $qd-rb$. But I don't know how to move from here. Can anyone point me in the right direction or tell me what I'm doing wrong?






abstract-algebra group-isomorphism quotient-group







share|cite|improve this question











$endgroup$












  • $begingroup$
    I'm assuming that definition of $B$ is a typo?
    $endgroup$
    – user3482749
    Jan 19 at 18:29










  • $begingroup$
    "That means that qc−ra should be equal to qd−rb." Not unless you think $G/H$ is trivial (which it clearly isn't).
    $endgroup$
    – user3482749
    Jan 19 at 18:30










  • $begingroup$
    Yes B was a typo, edited the question
    $endgroup$
    – Mathis
    Jan 19 at 18:32










  • $begingroup$
    @user3482749 "Not unless you think G/H is trivial" can you explain?
    $endgroup$
    – Mathis
    Jan 19 at 18:55










  • $begingroup$
    If you could prove, just from what you've got so far, that $qc - ra = qd - rb$, then you would have that $A^{-1}Bin H$ for all $A$ and $B$, hence $G/H$ would be trivial. That isn't the case, so that isn't going to be a useful line of enquiry.
    $endgroup$
    – user3482749
    Jan 19 at 19:08
















0












$begingroup$



Let $$G = left{begin{pmatrix} q&0\a+bi&qend{pmatrix} mid q in mathbb{Q}^ast, a,binmathbb{R}right}$$
and $$H = left{begin{pmatrix} q&0\a+ai&qend{pmatrix} mid q in mathbb{Q},q>0, ainmathbb{R}right}$$
Which known group is the quotient group $(G, ·)/H$ isomorphic to?




I don't know how to solve this. What I've tried:



$A,B in G$ belong to the same coset iff $A^{-1}B in H$.
Let $A = begin{pmatrix} q&0\a+bi&qend{pmatrix}$ and $B = begin{pmatrix} r&0\c+di&rend{pmatrix}$. Then $A^{-1} = begin{pmatrix} 1/q&0\(-a-bi)/q^2&1/qend{pmatrix}$ and $A^{-1}B = begin{pmatrix} r/q&0\frac{qc-ra+i(qd-rb)}{q^2}&r/qend{pmatrix}$



That means that $qc-ra$ should be equal to $qd-rb$. But I don't know how to move from here. Can anyone point me in the right direction or tell me what I'm doing wrong?






abstract-algebra group-isomorphism quotient-group







share|cite|improve this question











$endgroup$












  • $begingroup$
    I'm assuming that definition of $B$ is a typo?
    $endgroup$
    – user3482749
    Jan 19 at 18:29










  • $begingroup$
    "That means that qc−ra should be equal to qd−rb." Not unless you think $G/H$ is trivial (which it clearly isn't).
    $endgroup$
    – user3482749
    Jan 19 at 18:30










  • $begingroup$
    Yes B was a typo, edited the question
    $endgroup$
    – Mathis
    Jan 19 at 18:32










  • $begingroup$
    @user3482749 "Not unless you think G/H is trivial" can you explain?
    $endgroup$
    – Mathis
    Jan 19 at 18:55










  • $begingroup$
    If you could prove, just from what you've got so far, that $qc - ra = qd - rb$, then you would have that $A^{-1}Bin H$ for all $A$ and $B$, hence $G/H$ would be trivial. That isn't the case, so that isn't going to be a useful line of enquiry.
    $endgroup$
    – user3482749
    Jan 19 at 19:08














0












0








0


1



$begingroup$



Let $$G = left{begin{pmatrix} q&0\a+bi&qend{pmatrix} mid q in mathbb{Q}^ast, a,binmathbb{R}right}$$
and $$H = left{begin{pmatrix} q&0\a+ai&qend{pmatrix} mid q in mathbb{Q},q>0, ainmathbb{R}right}$$
Which known group is the quotient group $(G, ·)/H$ isomorphic to?




I don't know how to solve this. What I've tried:



$A,B in G$ belong to the same coset iff $A^{-1}B in H$.
Let $A = begin{pmatrix} q&0\a+bi&qend{pmatrix}$ and $B = begin{pmatrix} r&0\c+di&rend{pmatrix}$. Then $A^{-1} = begin{pmatrix} 1/q&0\(-a-bi)/q^2&1/qend{pmatrix}$ and $A^{-1}B = begin{pmatrix} r/q&0\frac{qc-ra+i(qd-rb)}{q^2}&r/qend{pmatrix}$



That means that $qc-ra$ should be equal to $qd-rb$. But I don't know how to move from here. Can anyone point me in the right direction or tell me what I'm doing wrong?






abstract-algebra group-isomorphism quotient-group







share|cite|improve this question











$endgroup$





Let $$G = left{begin{pmatrix} q&0\a+bi&qend{pmatrix} mid q in mathbb{Q}^ast, a,binmathbb{R}right}$$
and $$H = left{begin{pmatrix} q&0\a+ai&qend{pmatrix} mid q in mathbb{Q},q>0, ainmathbb{R}right}$$
Which known group is the quotient group $(G, ·)/H$ isomorphic to?




I don't know how to solve this. What I've tried:



$A,B in G$ belong to the same coset iff $A^{-1}B in H$.
Let $A = begin{pmatrix} q&0\a+bi&qend{pmatrix}$ and $B = begin{pmatrix} r&0\c+di&rend{pmatrix}$. Then $A^{-1} = begin{pmatrix} 1/q&0\(-a-bi)/q^2&1/qend{pmatrix}$ and $A^{-1}B = begin{pmatrix} r/q&0\frac{qc-ra+i(qd-rb)}{q^2}&r/qend{pmatrix}$



That means that $qc-ra$ should be equal to $qd-rb$. But I don't know how to move from here. Can anyone point me in the right direction or tell me what I'm doing wrong?






abstract-algebra group-isomorphism quotient-group




abstract-algebra group-isomorphism quotient-group






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 19 at 20:25









Henrik

6,03792030




6,03792030










asked Jan 19 at 18:26









MathisMathis

153




153












  • $begingroup$
    I'm assuming that definition of $B$ is a typo?
    $endgroup$
    – user3482749
    Jan 19 at 18:29










  • $begingroup$
    "That means that qc−ra should be equal to qd−rb." Not unless you think $G/H$ is trivial (which it clearly isn't).
    $endgroup$
    – user3482749
    Jan 19 at 18:30










  • $begingroup$
    Yes B was a typo, edited the question
    $endgroup$
    – Mathis
    Jan 19 at 18:32










  • $begingroup$
    @user3482749 "Not unless you think G/H is trivial" can you explain?
    $endgroup$
    – Mathis
    Jan 19 at 18:55










  • $begingroup$
    If you could prove, just from what you've got so far, that $qc - ra = qd - rb$, then you would have that $A^{-1}Bin H$ for all $A$ and $B$, hence $G/H$ would be trivial. That isn't the case, so that isn't going to be a useful line of enquiry.
    $endgroup$
    – user3482749
    Jan 19 at 19:08


















  • $begingroup$
    I'm assuming that definition of $B$ is a typo?
    $endgroup$
    – user3482749
    Jan 19 at 18:29










  • $begingroup$
    "That means that qc−ra should be equal to qd−rb." Not unless you think $G/H$ is trivial (which it clearly isn't).
    $endgroup$
    – user3482749
    Jan 19 at 18:30










  • $begingroup$
    Yes B was a typo, edited the question
    $endgroup$
    – Mathis
    Jan 19 at 18:32










  • $begingroup$
    @user3482749 "Not unless you think G/H is trivial" can you explain?
    $endgroup$
    – Mathis
    Jan 19 at 18:55










  • $begingroup$
    If you could prove, just from what you've got so far, that $qc - ra = qd - rb$, then you would have that $A^{-1}Bin H$ for all $A$ and $B$, hence $G/H$ would be trivial. That isn't the case, so that isn't going to be a useful line of enquiry.
    $endgroup$
    – user3482749
    Jan 19 at 19:08
















$begingroup$
I'm assuming that definition of $B$ is a typo?
$endgroup$
– user3482749
Jan 19 at 18:29




$begingroup$
I'm assuming that definition of $B$ is a typo?
$endgroup$
– user3482749
Jan 19 at 18:29












$begingroup$
"That means that qc−ra should be equal to qd−rb." Not unless you think $G/H$ is trivial (which it clearly isn't).
$endgroup$
– user3482749
Jan 19 at 18:30




$begingroup$
"That means that qc−ra should be equal to qd−rb." Not unless you think $G/H$ is trivial (which it clearly isn't).
$endgroup$
– user3482749
Jan 19 at 18:30












$begingroup$
Yes B was a typo, edited the question
$endgroup$
– Mathis
Jan 19 at 18:32




$begingroup$
Yes B was a typo, edited the question
$endgroup$
– Mathis
Jan 19 at 18:32












$begingroup$
@user3482749 "Not unless you think G/H is trivial" can you explain?
$endgroup$
– Mathis
Jan 19 at 18:55




$begingroup$
@user3482749 "Not unless you think G/H is trivial" can you explain?
$endgroup$
– Mathis
Jan 19 at 18:55












$begingroup$
If you could prove, just from what you've got so far, that $qc - ra = qd - rb$, then you would have that $A^{-1}Bin H$ for all $A$ and $B$, hence $G/H$ would be trivial. That isn't the case, so that isn't going to be a useful line of enquiry.
$endgroup$
– user3482749
Jan 19 at 19:08




$begingroup$
If you could prove, just from what you've got so far, that $qc - ra = qd - rb$, then you would have that $A^{-1}Bin H$ for all $A$ and $B$, hence $G/H$ would be trivial. That isn't the case, so that isn't going to be a useful line of enquiry.
$endgroup$
– user3482749
Jan 19 at 19:08










1 Answer
1






active

oldest

votes


















1












$begingroup$

Clearly we can identify $(G,cdot)$ with $(mathbb{Q}^* times mathbb{C},*)$ by



$$
begin{pmatrix}
q & 0 \
z & q
end{pmatrix}
mapsto (q,z)
$$
where $(q,z)*(r,w) = (qr, qw+rz)$. Note that $(mathbb{Q}^* times mathbb{C},*) simeq (mathbb{Q}^*,cdot) oplus (mathbb{C},+)$, in fact $(q,z) = (q,0)*(1,frac{z}{q})$ uniquely and



$$ ({ (q,0) in G: q in mathbb{Q}^* },*) simeq (mathbb{Q}^*,cdot) $$
$$ ({ (1,z) in G: z in mathbb{C} },*) simeq (mathbb{C},+) $$
since $(q,0)*(r,0) = (qr,0)$ and $(1,z)*(1,w) = (1,z+w)$. With this identification



$$ H simeq (mathbb{Q}^+,cdot) oplus (mathbb{R}(1+i),+) simeq (mathbb{Q}^+,cdot) oplus(mathbb{R},+). $$
Now since
$$ a+ib = frac{1}{2}Big( (a+b)(1+i)+(a-b)(1-i) Big), $$
we have $mathbb{C} simeq mathbb{R}(1+i) oplus mathbb{R}(1-i) simeq mathbb{R}oplus mathbb{R}$. Finally



$$ frac{G}{H} simeq frac{(mathbb{Q}^*,cdot) oplus (mathbb{C},+)}{(mathbb{Q}^+,cdot) oplus (mathbb{R},+)} simeq ({pm1} times mathbb{R},*). $$
Now $({pm1} times mathbb{R},*) simeq (mathbb{R},+)$ by $varphi$ defined $(a,r) mapsto ar$, in fact
$$ varphi Big( (a,r)*(b,s) Big) = varphi(ab,as+br) = ab(as+br) = bs + ar = varphi(a,r)+varphi(b,s). $$



Therefore $(G/H,cdot) simeq (mathbb{R},+)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for the answer, but I have a few questions. What operation is ⊕? How do you get from ab(as+br) to bs+ar?
    $endgroup$
    – Mathis
    Jan 20 at 10:24










  • $begingroup$
    The symbol $oplus$ denote the direct sum of groups en.wikipedia.org/wiki/Direct_sum. If $a in {pm 1}$, then $a^2 =1$.
    $endgroup$
    – Kalhac
    Jan 20 at 14:25













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Clearly we can identify $(G,cdot)$ with $(mathbb{Q}^* times mathbb{C},*)$ by



$$
begin{pmatrix}
q & 0 \
z & q
end{pmatrix}
mapsto (q,z)
$$
where $(q,z)*(r,w) = (qr, qw+rz)$. Note that $(mathbb{Q}^* times mathbb{C},*) simeq (mathbb{Q}^*,cdot) oplus (mathbb{C},+)$, in fact $(q,z) = (q,0)*(1,frac{z}{q})$ uniquely and



$$ ({ (q,0) in G: q in mathbb{Q}^* },*) simeq (mathbb{Q}^*,cdot) $$
$$ ({ (1,z) in G: z in mathbb{C} },*) simeq (mathbb{C},+) $$
since $(q,0)*(r,0) = (qr,0)$ and $(1,z)*(1,w) = (1,z+w)$. With this identification



$$ H simeq (mathbb{Q}^+,cdot) oplus (mathbb{R}(1+i),+) simeq (mathbb{Q}^+,cdot) oplus(mathbb{R},+). $$
Now since
$$ a+ib = frac{1}{2}Big( (a+b)(1+i)+(a-b)(1-i) Big), $$
we have $mathbb{C} simeq mathbb{R}(1+i) oplus mathbb{R}(1-i) simeq mathbb{R}oplus mathbb{R}$. Finally



$$ frac{G}{H} simeq frac{(mathbb{Q}^*,cdot) oplus (mathbb{C},+)}{(mathbb{Q}^+,cdot) oplus (mathbb{R},+)} simeq ({pm1} times mathbb{R},*). $$
Now $({pm1} times mathbb{R},*) simeq (mathbb{R},+)$ by $varphi$ defined $(a,r) mapsto ar$, in fact
$$ varphi Big( (a,r)*(b,s) Big) = varphi(ab,as+br) = ab(as+br) = bs + ar = varphi(a,r)+varphi(b,s). $$



Therefore $(G/H,cdot) simeq (mathbb{R},+)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for the answer, but I have a few questions. What operation is ⊕? How do you get from ab(as+br) to bs+ar?
    $endgroup$
    – Mathis
    Jan 20 at 10:24










  • $begingroup$
    The symbol $oplus$ denote the direct sum of groups en.wikipedia.org/wiki/Direct_sum. If $a in {pm 1}$, then $a^2 =1$.
    $endgroup$
    – Kalhac
    Jan 20 at 14:25


















1












$begingroup$

Clearly we can identify $(G,cdot)$ with $(mathbb{Q}^* times mathbb{C},*)$ by



$$
begin{pmatrix}
q & 0 \
z & q
end{pmatrix}
mapsto (q,z)
$$
where $(q,z)*(r,w) = (qr, qw+rz)$. Note that $(mathbb{Q}^* times mathbb{C},*) simeq (mathbb{Q}^*,cdot) oplus (mathbb{C},+)$, in fact $(q,z) = (q,0)*(1,frac{z}{q})$ uniquely and



$$ ({ (q,0) in G: q in mathbb{Q}^* },*) simeq (mathbb{Q}^*,cdot) $$
$$ ({ (1,z) in G: z in mathbb{C} },*) simeq (mathbb{C},+) $$
since $(q,0)*(r,0) = (qr,0)$ and $(1,z)*(1,w) = (1,z+w)$. With this identification



$$ H simeq (mathbb{Q}^+,cdot) oplus (mathbb{R}(1+i),+) simeq (mathbb{Q}^+,cdot) oplus(mathbb{R},+). $$
Now since
$$ a+ib = frac{1}{2}Big( (a+b)(1+i)+(a-b)(1-i) Big), $$
we have $mathbb{C} simeq mathbb{R}(1+i) oplus mathbb{R}(1-i) simeq mathbb{R}oplus mathbb{R}$. Finally



$$ frac{G}{H} simeq frac{(mathbb{Q}^*,cdot) oplus (mathbb{C},+)}{(mathbb{Q}^+,cdot) oplus (mathbb{R},+)} simeq ({pm1} times mathbb{R},*). $$
Now $({pm1} times mathbb{R},*) simeq (mathbb{R},+)$ by $varphi$ defined $(a,r) mapsto ar$, in fact
$$ varphi Big( (a,r)*(b,s) Big) = varphi(ab,as+br) = ab(as+br) = bs + ar = varphi(a,r)+varphi(b,s). $$



Therefore $(G/H,cdot) simeq (mathbb{R},+)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for the answer, but I have a few questions. What operation is ⊕? How do you get from ab(as+br) to bs+ar?
    $endgroup$
    – Mathis
    Jan 20 at 10:24










  • $begingroup$
    The symbol $oplus$ denote the direct sum of groups en.wikipedia.org/wiki/Direct_sum. If $a in {pm 1}$, then $a^2 =1$.
    $endgroup$
    – Kalhac
    Jan 20 at 14:25
















1












1








1





$begingroup$

Clearly we can identify $(G,cdot)$ with $(mathbb{Q}^* times mathbb{C},*)$ by



$$
begin{pmatrix}
q & 0 \
z & q
end{pmatrix}
mapsto (q,z)
$$
where $(q,z)*(r,w) = (qr, qw+rz)$. Note that $(mathbb{Q}^* times mathbb{C},*) simeq (mathbb{Q}^*,cdot) oplus (mathbb{C},+)$, in fact $(q,z) = (q,0)*(1,frac{z}{q})$ uniquely and



$$ ({ (q,0) in G: q in mathbb{Q}^* },*) simeq (mathbb{Q}^*,cdot) $$
$$ ({ (1,z) in G: z in mathbb{C} },*) simeq (mathbb{C},+) $$
since $(q,0)*(r,0) = (qr,0)$ and $(1,z)*(1,w) = (1,z+w)$. With this identification



$$ H simeq (mathbb{Q}^+,cdot) oplus (mathbb{R}(1+i),+) simeq (mathbb{Q}^+,cdot) oplus(mathbb{R},+). $$
Now since
$$ a+ib = frac{1}{2}Big( (a+b)(1+i)+(a-b)(1-i) Big), $$
we have $mathbb{C} simeq mathbb{R}(1+i) oplus mathbb{R}(1-i) simeq mathbb{R}oplus mathbb{R}$. Finally



$$ frac{G}{H} simeq frac{(mathbb{Q}^*,cdot) oplus (mathbb{C},+)}{(mathbb{Q}^+,cdot) oplus (mathbb{R},+)} simeq ({pm1} times mathbb{R},*). $$
Now $({pm1} times mathbb{R},*) simeq (mathbb{R},+)$ by $varphi$ defined $(a,r) mapsto ar$, in fact
$$ varphi Big( (a,r)*(b,s) Big) = varphi(ab,as+br) = ab(as+br) = bs + ar = varphi(a,r)+varphi(b,s). $$



Therefore $(G/H,cdot) simeq (mathbb{R},+)$.






share|cite|improve this answer









$endgroup$



Clearly we can identify $(G,cdot)$ with $(mathbb{Q}^* times mathbb{C},*)$ by



$$
begin{pmatrix}
q & 0 \
z & q
end{pmatrix}
mapsto (q,z)
$$
where $(q,z)*(r,w) = (qr, qw+rz)$. Note that $(mathbb{Q}^* times mathbb{C},*) simeq (mathbb{Q}^*,cdot) oplus (mathbb{C},+)$, in fact $(q,z) = (q,0)*(1,frac{z}{q})$ uniquely and



$$ ({ (q,0) in G: q in mathbb{Q}^* },*) simeq (mathbb{Q}^*,cdot) $$
$$ ({ (1,z) in G: z in mathbb{C} },*) simeq (mathbb{C},+) $$
since $(q,0)*(r,0) = (qr,0)$ and $(1,z)*(1,w) = (1,z+w)$. With this identification



$$ H simeq (mathbb{Q}^+,cdot) oplus (mathbb{R}(1+i),+) simeq (mathbb{Q}^+,cdot) oplus(mathbb{R},+). $$
Now since
$$ a+ib = frac{1}{2}Big( (a+b)(1+i)+(a-b)(1-i) Big), $$
we have $mathbb{C} simeq mathbb{R}(1+i) oplus mathbb{R}(1-i) simeq mathbb{R}oplus mathbb{R}$. Finally



$$ frac{G}{H} simeq frac{(mathbb{Q}^*,cdot) oplus (mathbb{C},+)}{(mathbb{Q}^+,cdot) oplus (mathbb{R},+)} simeq ({pm1} times mathbb{R},*). $$
Now $({pm1} times mathbb{R},*) simeq (mathbb{R},+)$ by $varphi$ defined $(a,r) mapsto ar$, in fact
$$ varphi Big( (a,r)*(b,s) Big) = varphi(ab,as+br) = ab(as+br) = bs + ar = varphi(a,r)+varphi(b,s). $$



Therefore $(G/H,cdot) simeq (mathbb{R},+)$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 19 at 20:01









KalhacKalhac

33417




33417












  • $begingroup$
    Thank you for the answer, but I have a few questions. What operation is ⊕? How do you get from ab(as+br) to bs+ar?
    $endgroup$
    – Mathis
    Jan 20 at 10:24










  • $begingroup$
    The symbol $oplus$ denote the direct sum of groups en.wikipedia.org/wiki/Direct_sum. If $a in {pm 1}$, then $a^2 =1$.
    $endgroup$
    – Kalhac
    Jan 20 at 14:25




















  • $begingroup$
    Thank you for the answer, but I have a few questions. What operation is ⊕? How do you get from ab(as+br) to bs+ar?
    $endgroup$
    – Mathis
    Jan 20 at 10:24










  • $begingroup$
    The symbol $oplus$ denote the direct sum of groups en.wikipedia.org/wiki/Direct_sum. If $a in {pm 1}$, then $a^2 =1$.
    $endgroup$
    – Kalhac
    Jan 20 at 14:25


















$begingroup$
Thank you for the answer, but I have a few questions. What operation is ⊕? How do you get from ab(as+br) to bs+ar?
$endgroup$
– Mathis
Jan 20 at 10:24




$begingroup$
Thank you for the answer, but I have a few questions. What operation is ⊕? How do you get from ab(as+br) to bs+ar?
$endgroup$
– Mathis
Jan 20 at 10:24












$begingroup$
The symbol $oplus$ denote the direct sum of groups en.wikipedia.org/wiki/Direct_sum. If $a in {pm 1}$, then $a^2 =1$.
$endgroup$
– Kalhac
Jan 20 at 14:25






$begingroup$
The symbol $oplus$ denote the direct sum of groups en.wikipedia.org/wiki/Direct_sum. If $a in {pm 1}$, then $a^2 =1$.
$endgroup$
– Kalhac
Jan 20 at 14:25




















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