If $R'$ is set of all reals except zero and $a*b = |a|b$, then is $(R',*)$ a group?

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0
down vote

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Let $R'$ be set of all real numbers except zero. Define a binary operation $*$
such that $a*b = |a|b$, where $|a|$ denotes the absolute value of $a$. Does $(R'*)$ form a group?

My attempt:

1 is the identity. But here $-2*1 = 2$

So can we say it is not a group?

edited Jul 17 at 15:38

Arnaud Mortier

19.1k22159

asked Jul 17 at 15:25

Magneto

791213

4

"1 is identity. But here $-2*1 = 2$" But then $1$ isn't idrentity, since you've just found a counterexample.
â€“Â Arthur
Jul 17 at 15:29

add a commentÂ |Â

up vote
0
down vote

favorite

Let $R'$ be set of all real numbers except zero. Define a binary operation $*$
such that $a*b = |a|b$, where $|a|$ denotes the absolute value of $a$. Does $(R'*)$ form a group?

My attempt:

1 is the identity. But here $-2*1 = 2$

So can we say it is not a group?

edited Jul 17 at 15:38

Arnaud Mortier

19.1k22159

asked Jul 17 at 15:25

Magneto

791213

4

"1 is identity. But here $-2*1 = 2$" But then $1$ isn't idrentity, since you've just found a counterexample.
â€“Â Arthur
Jul 17 at 15:29

add a commentÂ |Â

up vote
0
down vote

favorite

Let $R'$ be set of all real numbers except zero. Define a binary operation $*$
such that $a*b = |a|b$, where $|a|$ denotes the absolute value of $a$. Does $(R'*)$ form a group?

My attempt:

1 is the identity. But here $-2*1 = 2$

So can we say it is not a group?

edited Jul 17 at 15:38

Arnaud Mortier

19.1k22159

asked Jul 17 at 15:25

Magneto

791213

Let $R'$ be set of all real numbers except zero. Define a binary operation $*$
such that $a*b = |a|b$, where $|a|$ denotes the absolute value of $a$. Does $(R'*)$ form a group?

My attempt:

1 is the identity. But here $-2*1 = 2$

So can we say it is not a group?

edited Jul 17 at 15:38

Arnaud Mortier

19.1k22159

asked Jul 17 at 15:25

Magneto

791213

edited Jul 17 at 15:38

Arnaud Mortier

19.1k22159

edited Jul 17 at 15:38

Arnaud Mortier

19.1k22159

edited Jul 17 at 15:38

Arnaud Mortier

19.1k22159

asked Jul 17 at 15:25

Magneto

791213

asked Jul 17 at 15:25

Magneto

791213

asked Jul 17 at 15:25

Magneto

791213

4

"1 is identity. But here $-2*1 = 2$" But then $1$ isn't idrentity, since you've just found a counterexample.
â€“Â Arthur
Jul 17 at 15:29

add a commentÂ |Â

4

"1 is identity. But here $-2*1 = 2$" But then $1$ isn't idrentity, since you've just found a counterexample.
â€“Â Arthur
Jul 17 at 15:29

"1 is identity. But here $-2*1 = 2$" But then $1$ isn't idrentity, since you've just found a counterexample.
â€“Â Arthur
Jul 17 at 15:29

add a commentÂ |Â

2 Answers
2

active

oldest

votes

up vote
4
down vote

accepted

1 is the identity of the real numbers under standard multiplication, but since you are not using standard multiplication, then you could imagine that some other real number $lambda$ could be the identity. However, that number would need to satisfy:

$-2 * lambda = -2$ and $2 * lambda = 2$.

but then, $2lambda = -2$ and $2lambda = 2$. Clearly, no real number satisfies both of these properties, so it is not a group.

answered Jul 17 at 15:33

user145640

1665

+1. Note to the OP that merely saying "$1$ looks like it should be the identity, but it's not" isn't enough to conclude that the structure isn't a group; it's always possible that some other element could be the identity. To show that it's not a group, you need to show that no element at all is the identity, which is what this answer does.
â€“Â Noah Schweber
Jul 17 at 15:35

thank you sir, got it
â€“Â Magneto
Jul 18 at 3:46

add a commentÂ |Â

up vote
4
down vote

Mostly.

The identity is not $1$ but you have to prove that it can't be anything else.

Let $e in R'$

If $e > 0$ let $a in R'; a < 0$ then $|a|e > 0$ so $|a|e$ is not the identity.

If $e < 0$ let $a in R'; a > 0$ then $|a|e < 0$ so $|a|e$ is not the identity.

So it is not possible for any arbitrary $e$ to be the identity.

So there is no identity.

====

Alternatively, to give the OP a little more credit (perhaps more than is actually there)

One could say:

Suppose there were an identity.

Then $|2|*e = 2$ so that would mean $e= 1$ and therefore $1$ is the only possible candidate to be the identity.

But $|-2|*1 ne -2$. So it is not the identity. So there is not identity.

To claim "$1$ is the identity", which I'm leniently going to accept as meaning "$1$ should be the identity", one does have to explain why one thinks $1$ should be the identity.

edited Jul 17 at 15:44

answered Jul 17 at 15:37

fleablood

60.5k22575

very elaborative. thank you
â€“Â Magneto
Jul 18 at 3:46

add a commentÂ |Â

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
4
down vote

accepted

$-2 * lambda = -2$ and $2 * lambda = 2$.

but then, $2lambda = -2$ and $2lambda = 2$. Clearly, no real number satisfies both of these properties, so it is not a group.

answered Jul 17 at 15:33

user145640

1665

+1. Note to the OP that merely saying "$1$ looks like it should be the identity, but it's not" isn't enough to conclude that the structure isn't a group; it's always possible that some other element could be the identity. To show that it's not a group, you need to show that no element at all is the identity, which is what this answer does.
â€“Â Noah Schweber
Jul 17 at 15:35

thank you sir, got it
â€“Â Magneto
Jul 18 at 3:46

add a commentÂ |Â

up vote
4
down vote

accepted

$-2 * lambda = -2$ and $2 * lambda = 2$.

but then, $2lambda = -2$ and $2lambda = 2$. Clearly, no real number satisfies both of these properties, so it is not a group.

answered Jul 17 at 15:33

user145640

1665

+1. Note to the OP that merely saying "$1$ looks like it should be the identity, but it's not" isn't enough to conclude that the structure isn't a group; it's always possible that some other element could be the identity. To show that it's not a group, you need to show that no element at all is the identity, which is what this answer does.
â€“Â Noah Schweber
Jul 17 at 15:35

thank you sir, got it
â€“Â Magneto
Jul 18 at 3:46

add a commentÂ |Â

up vote
4
down vote

accepted

$-2 * lambda = -2$ and $2 * lambda = 2$.

but then, $2lambda = -2$ and $2lambda = 2$. Clearly, no real number satisfies both of these properties, so it is not a group.

answered Jul 17 at 15:33

user145640

1665

$-2 * lambda = -2$ and $2 * lambda = 2$.

but then, $2lambda = -2$ and $2lambda = 2$. Clearly, no real number satisfies both of these properties, so it is not a group.

answered Jul 17 at 15:33

user145640

1665

answered Jul 17 at 15:33

user145640

1665

answered Jul 17 at 15:33

user145640

1665

answered Jul 17 at 15:33

user145640

1665

+1. Note to the OP that merely saying "$1$ looks like it should be the identity, but it's not" isn't enough to conclude that the structure isn't a group; it's always possible that some other element could be the identity. To show that it's not a group, you need to show that no element at all is the identity, which is what this answer does.
â€“Â Noah Schweber
Jul 17 at 15:35

thank you sir, got it
â€“Â Magneto
Jul 18 at 3:46

add a commentÂ |Â

+1. Note to the OP that merely saying "$1$ looks like it should be the identity, but it's not" isn't enough to conclude that the structure isn't a group; it's always possible that some other element could be the identity. To show that it's not a group, you need to show that no element at all is the identity, which is what this answer does.
â€“Â Noah Schweber
Jul 17 at 15:35

thank you sir, got it
â€“Â Magneto
Jul 18 at 3:46

+1. Note to the OP that merely saying "$1$ looks like it should be the identity, but it's not" isn't enough to conclude that the structure isn't a group; it's always possible that some other element could be the identity. To show that it's not a group, you need to show that no element at all is the identity, which is what this answer does.
â€“Â Noah Schweber
Jul 17 at 15:35

thank you sir, got it
â€“Â Magneto
Jul 18 at 3:46

add a commentÂ |Â

up vote
4
down vote

Mostly.

The identity is not $1$ but you have to prove that it can't be anything else.

Let $e in R'$

If $e > 0$ let $a in R'; a < 0$ then $|a|e > 0$ so $|a|e$ is not the identity.

If $e < 0$ let $a in R'; a > 0$ then $|a|e < 0$ so $|a|e$ is not the identity.

So it is not possible for any arbitrary $e$ to be the identity.

So there is no identity.

====

Alternatively, to give the OP a little more credit (perhaps more than is actually there)

One could say:

Suppose there were an identity.

Then $|2|*e = 2$ so that would mean $e= 1$ and therefore $1$ is the only possible candidate to be the identity.

But $|-2|*1 ne -2$. So it is not the identity. So there is not identity.

To claim "$1$ is the identity", which I'm leniently going to accept as meaning "$1$ should be the identity", one does have to explain why one thinks $1$ should be the identity.

edited Jul 17 at 15:44

answered Jul 17 at 15:37

fleablood

60.5k22575

very elaborative. thank you
â€“Â Magneto
Jul 18 at 3:46

add a commentÂ |Â

up vote
4
down vote

Mostly.

The identity is not $1$ but you have to prove that it can't be anything else.

Let $e in R'$

If $e > 0$ let $a in R'; a < 0$ then $|a|e > 0$ so $|a|e$ is not the identity.

If $e < 0$ let $a in R'; a > 0$ then $|a|e < 0$ so $|a|e$ is not the identity.

So it is not possible for any arbitrary $e$ to be the identity.

So there is no identity.

====

Alternatively, to give the OP a little more credit (perhaps more than is actually there)

One could say:

Suppose there were an identity.

Then $|2|*e = 2$ so that would mean $e= 1$ and therefore $1$ is the only possible candidate to be the identity.

But $|-2|*1 ne -2$. So it is not the identity. So there is not identity.

To claim "$1$ is the identity", which I'm leniently going to accept as meaning "$1$ should be the identity", one does have to explain why one thinks $1$ should be the identity.

edited Jul 17 at 15:44

answered Jul 17 at 15:37

fleablood

60.5k22575

very elaborative. thank you
â€“Â Magneto
Jul 18 at 3:46

add a commentÂ |Â

up vote
4
down vote

Mostly.

The identity is not $1$ but you have to prove that it can't be anything else.

Let $e in R'$

If $e > 0$ let $a in R'; a < 0$ then $|a|e > 0$ so $|a|e$ is not the identity.

If $e < 0$ let $a in R'; a > 0$ then $|a|e < 0$ so $|a|e$ is not the identity.

So it is not possible for any arbitrary $e$ to be the identity.

So there is no identity.

====

Alternatively, to give the OP a little more credit (perhaps more than is actually there)

One could say:

Suppose there were an identity.

Then $|2|*e = 2$ so that would mean $e= 1$ and therefore $1$ is the only possible candidate to be the identity.

But $|-2|*1 ne -2$. So it is not the identity. So there is not identity.

To claim "$1$ is the identity", which I'm leniently going to accept as meaning "$1$ should be the identity", one does have to explain why one thinks $1$ should be the identity.

edited Jul 17 at 15:44

answered Jul 17 at 15:37

fleablood

60.5k22575

Mostly.

The identity is not $1$ but you have to prove that it can't be anything else.

Let $e in R'$

If $e > 0$ let $a in R'; a < 0$ then $|a|e > 0$ so $|a|e$ is not the identity.

If $e < 0$ let $a in R'; a > 0$ then $|a|e < 0$ so $|a|e$ is not the identity.

So it is not possible for any arbitrary $e$ to be the identity.

So there is no identity.

====

Alternatively, to give the OP a little more credit (perhaps more than is actually there)

One could say:

Suppose there were an identity.

Then $|2|*e = 2$ so that would mean $e= 1$ and therefore $1$ is the only possible candidate to be the identity.

But $|-2|*1 ne -2$. So it is not the identity. So there is not identity.

To claim "$1$ is the identity", which I'm leniently going to accept as meaning "$1$ should be the identity", one does have to explain why one thinks $1$ should be the identity.

edited Jul 17 at 15:44

answered Jul 17 at 15:37

fleablood

60.5k22575

edited Jul 17 at 15:44

answered Jul 17 at 15:37

fleablood

60.5k22575

answered Jul 17 at 15:37

fleablood

60.5k22575

answered Jul 17 at 15:37

fleablood

60.5k22575

very elaborative. thank you
â€“Â Magneto
Jul 18 at 3:46

add a commentÂ |Â

very elaborative. thank you
â€“Â Magneto
Jul 18 at 3:46

very elaborative. thank you
â€“Â Magneto
Jul 18 at 3:46

add a commentÂ |Â

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