Can elements in a group have more than one inverse
$begingroup$
I had to prove if a set of $2times 2$ matrices is a group with respect to multiplication where I define a group as the set of all
$$A_alpha = begin{bmatrix}cosalpha&-sinalpha \ sinalpha & cosalpha end{bmatrix}$$
where $alphainBbb R$.
Now its fairly easy to prove that this set is non-empty and satisfies closure and associativity w.r.t multiplication as well as identity. What I was confused about here was that since $A_{2pi+alpha}=A_alpha$, then will this set be considered to have more than one identity element as $A_{2npi}$ will be the identity element.
Likewise, when finding out the inverse, I had already derived that $A_alpha*A_beta=A_{alpha+beta}$, if I took inverse of $A_alpha$ to be $A_beta$, the relation between $beta;text{and};alpha;text{is}$ $$beta=2npi-alpha$$
Here, again, although we get the same matrices for all the different $beta$ values, will we consider $alpha$ to have more than one inverses.
In short, what I am confused about is that we get a single matrix for a lot of different values of $alpha$. So are they all supposed to be considered as the same element of the set of matrices.
matrices group-theory trigonometry elementary-set-theory
edited Jan 4 '18 at 9:56
Arnaud D.
16.2k52444
asked Jan 4 '18 at 9:49
Siddharth JhaSiddharth Jha
589
$endgroup$
add a comment |
$begingroup$
I had to prove if a set of $2times 2$ matrices is a group with respect to multiplication where I define a group as the set of all
$$A_alpha = begin{bmatrix}cosalpha&-sinalpha \ sinalpha & cosalpha end{bmatrix}$$
where $alphainBbb R$.
Now its fairly easy to prove that this set is non-empty and satisfies closure and associativity w.r.t multiplication as well as identity. What I was confused about here was that since $A_{2pi+alpha}=A_alpha$, then will this set be considered to have more than one identity element as $A_{2npi}$ will be the identity element.
Likewise, when finding out the inverse, I had already derived that $A_alpha*A_beta=A_{alpha+beta}$, if I took inverse of $A_alpha$ to be $A_beta$, the relation between $beta;text{and};alpha;text{is}$ $$beta=2npi-alpha$$
Here, again, although we get the same matrices for all the different $beta$ values, will we consider $alpha$ to have more than one inverses.
In short, what I am confused about is that we get a single matrix for a lot of different values of $alpha$. So are they all supposed to be considered as the same element of the set of matrices.
matrices group-theory trigonometry elementary-set-theory
edited Jan 4 '18 at 9:56
Arnaud D.
16.2k52444
asked Jan 4 '18 at 9:49
Siddharth JhaSiddharth Jha
589
$endgroup$
add a comment |
$begingroup$
I had to prove if a set of $2times 2$ matrices is a group with respect to multiplication where I define a group as the set of all
$$A_alpha = begin{bmatrix}cosalpha&-sinalpha \ sinalpha & cosalpha end{bmatrix}$$
where $alphainBbb R$.
Now its fairly easy to prove that this set is non-empty and satisfies closure and associativity w.r.t multiplication as well as identity. What I was confused about here was that since $A_{2pi+alpha}=A_alpha$, then will this set be considered to have more than one identity element as $A_{2npi}$ will be the identity element.
Likewise, when finding out the inverse, I had already derived that $A_alpha*A_beta=A_{alpha+beta}$, if I took inverse of $A_alpha$ to be $A_beta$, the relation between $beta;text{and};alpha;text{is}$ $$beta=2npi-alpha$$
Here, again, although we get the same matrices for all the different $beta$ values, will we consider $alpha$ to have more than one inverses.
In short, what I am confused about is that we get a single matrix for a lot of different values of $alpha$. So are they all supposed to be considered as the same element of the set of matrices.
matrices group-theory trigonometry elementary-set-theory
edited Jan 4 '18 at 9:56
Arnaud D.
16.2k52444
asked Jan 4 '18 at 9:49
Siddharth JhaSiddharth Jha
589
$endgroup$
I had to prove if a set of $2times 2$ matrices is a group with respect to multiplication where I define a group as the set of all
$$A_alpha = begin{bmatrix}cosalpha&-sinalpha \ sinalpha & cosalpha end{bmatrix}$$
where $alphainBbb R$.
Now its fairly easy to prove that this set is non-empty and satisfies closure and associativity w.r.t multiplication as well as identity. What I was confused about here was that since $A_{2pi+alpha}=A_alpha$, then will this set be considered to have more than one identity element as $A_{2npi}$ will be the identity element.
Likewise, when finding out the inverse, I had already derived that $A_alpha*A_beta=A_{alpha+beta}$, if I took inverse of $A_alpha$ to be $A_beta$, the relation between $beta;text{and};alpha;text{is}$ $$beta=2npi-alpha$$
Here, again, although we get the same matrices for all the different $beta$ values, will we consider $alpha$ to have more than one inverses.
In short, what I am confused about is that we get a single matrix for a lot of different values of $alpha$. So are they all supposed to be considered as the same element of the set of matrices.
matrices group-theory trigonometry elementary-set-theory
matrices group-theory trigonometry elementary-set-theory
edited Jan 4 '18 at 9:56
Arnaud D.
16.2k52444
asked Jan 4 '18 at 9:49
Siddharth JhaSiddharth Jha
589
edited Jan 4 '18 at 9:56
Arnaud D.
16.2k52444
asked Jan 4 '18 at 9:49
Siddharth JhaSiddharth Jha
589
edited Jan 4 '18 at 9:56
Arnaud D.
16.2k52444
edited Jan 4 '18 at 9:56
Arnaud D.
16.2k52444
edited Jan 4 '18 at 9:56
Arnaud D.
16.2k52444
16.2k52444
asked Jan 4 '18 at 9:49
Siddharth JhaSiddharth Jha
589
asked Jan 4 '18 at 9:49
Siddharth JhaSiddharth Jha
589
asked Jan 4 '18 at 9:49
Siddharth JhaSiddharth Jha
589
589
add a comment |
add a comment |
2 Answers
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$begingroup$
"No" as answer on the question in the title.
If $a=b$ then the set ${a,b}$ has exactly one element.
You could say that $a$ and $b$ are just labels of the same mathematical object.
In your case ${A_{2npi+alpha}mid ninmathbb Z}={A_{alpha}}$ since $A_{2npi+alpha}=A_{alpha}$ for each $ninmathbb Z$.
So there is exactly one identity.
Same story for inverses.
answered Jan 4 '18 at 9:56
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
The catch here is that $A_{alpha+2pi}$ is actually equal to $A_{alpha}$ (since $cos{(x+2pi)}=cos(x)$ and likewise for $sin$). So the matrices $A_{-alpha}$, $A_{2pi-alpha}$ are indeed both inverses for $A_{alpha}$, but that is not a problem, since they are the same element!
answered Jan 4 '18 at 9:53
idokidok
1,405313
$endgroup$
add a comment |
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2 Answers
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2 Answers
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active
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$begingroup$
"No" as answer on the question in the title.
If $a=b$ then the set ${a,b}$ has exactly one element.
You could say that $a$ and $b$ are just labels of the same mathematical object.
In your case ${A_{2npi+alpha}mid ninmathbb Z}={A_{alpha}}$ since $A_{2npi+alpha}=A_{alpha}$ for each $ninmathbb Z$.
So there is exactly one identity.
Same story for inverses.
answered Jan 4 '18 at 9:56
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
"No" as answer on the question in the title.
If $a=b$ then the set ${a,b}$ has exactly one element.
You could say that $a$ and $b$ are just labels of the same mathematical object.
In your case ${A_{2npi+alpha}mid ninmathbb Z}={A_{alpha}}$ since $A_{2npi+alpha}=A_{alpha}$ for each $ninmathbb Z$.
So there is exactly one identity.
Same story for inverses.
answered Jan 4 '18 at 9:56
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
"No" as answer on the question in the title.
If $a=b$ then the set ${a,b}$ has exactly one element.
You could say that $a$ and $b$ are just labels of the same mathematical object.
In your case ${A_{2npi+alpha}mid ninmathbb Z}={A_{alpha}}$ since $A_{2npi+alpha}=A_{alpha}$ for each $ninmathbb Z$.
So there is exactly one identity.
Same story for inverses.
answered Jan 4 '18 at 9:56
drhabdrhab
103k545136
$endgroup$
"No" as answer on the question in the title.
If $a=b$ then the set ${a,b}$ has exactly one element.
You could say that $a$ and $b$ are just labels of the same mathematical object.
In your case ${A_{2npi+alpha}mid ninmathbb Z}={A_{alpha}}$ since $A_{2npi+alpha}=A_{alpha}$ for each $ninmathbb Z$.
So there is exactly one identity.
Same story for inverses.
answered Jan 4 '18 at 9:56
drhabdrhab
103k545136
edited Jan 4 '18 at 10:02
answered Jan 4 '18 at 9:56
drhabdrhab
103k545136
answered Jan 4 '18 at 9:56
drhabdrhab
103k545136
answered Jan 4 '18 at 9:56
drhabdrhab
103k545136
103k545136
add a comment |
add a comment |
$begingroup$
The catch here is that $A_{alpha+2pi}$ is actually equal to $A_{alpha}$ (since $cos{(x+2pi)}=cos(x)$ and likewise for $sin$). So the matrices $A_{-alpha}$, $A_{2pi-alpha}$ are indeed both inverses for $A_{alpha}$, but that is not a problem, since they are the same element!
answered Jan 4 '18 at 9:53
idokidok
1,405313
$endgroup$
add a comment |
$begingroup$
The catch here is that $A_{alpha+2pi}$ is actually equal to $A_{alpha}$ (since $cos{(x+2pi)}=cos(x)$ and likewise for $sin$). So the matrices $A_{-alpha}$, $A_{2pi-alpha}$ are indeed both inverses for $A_{alpha}$, but that is not a problem, since they are the same element!
answered Jan 4 '18 at 9:53
idokidok
1,405313
$endgroup$
add a comment |
$begingroup$
The catch here is that $A_{alpha+2pi}$ is actually equal to $A_{alpha}$ (since $cos{(x+2pi)}=cos(x)$ and likewise for $sin$). So the matrices $A_{-alpha}$, $A_{2pi-alpha}$ are indeed both inverses for $A_{alpha}$, but that is not a problem, since they are the same element!
answered Jan 4 '18 at 9:53
idokidok
1,405313
$endgroup$
The catch here is that $A_{alpha+2pi}$ is actually equal to $A_{alpha}$ (since $cos{(x+2pi)}=cos(x)$ and likewise for $sin$). So the matrices $A_{-alpha}$, $A_{2pi-alpha}$ are indeed both inverses for $A_{alpha}$, but that is not a problem, since they are the same element!
answered Jan 4 '18 at 9:53
idokidok
1,405313
answered Jan 4 '18 at 9:53
idokidok
1,405313
answered Jan 4 '18 at 9:53
idokidok
1,405313
answered Jan 4 '18 at 9:53
idokidok
1,405313
1,405313
add a comment |
add a comment |
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