Mixing
Is ${0,1}$ a subgroup of $mathbb{R}$ under multiplication?
$begingroup$
I am playing around with subgroups and I was wondering if the group ${0,1}$ is a subgroup of $mathbb{R}$ with respect to multiplication. It seems to fit the criteria but the inverse property is making me worry because $0$ is its own inverse. I vaguely remember a proof that if an element was its own inverse than it must be the identity element but $1$ would need to be the identity here. Thanks in advance!
abstract-algebra group-theory finite-groups
edited Jan 11 at 17:52
6005
36.2k751125
asked Jan 11 at 17:13
Ed W.Ed W.
112
$endgroup$
add a comment |
$begingroup$
I am playing around with subgroups and I was wondering if the group ${0,1}$ is a subgroup of $mathbb{R}$ with respect to multiplication. It seems to fit the criteria but the inverse property is making me worry because $0$ is its own inverse. I vaguely remember a proof that if an element was its own inverse than it must be the identity element but $1$ would need to be the identity here. Thanks in advance!
abstract-algebra group-theory finite-groups
edited Jan 11 at 17:52
6005
36.2k751125
asked Jan 11 at 17:13
Ed W.Ed W.
112
$endgroup$
2
$begingroup$
Assuming you mean the set ${0,1}subseteq mathbb{R}$ and $(mathbb{R},cdot)$ were a group under multiplication (which is not !). If $0$ were to have a multiplicative inverse then $exists; bin{0,1};:;0cdot b=1$. Is this possible? And if $a=a^{-1}$ then $a$ need not be identity. Consider the Klein-$4$ group ${1,-1,i,-i}$.
$endgroup$
– Yadati Kiran
Jan 11 at 17:16
$begingroup$
I see, I definitely need to do some more brushing up on this, thank you!
$endgroup$
– Ed W.
Jan 11 at 17:26
$begingroup$
I am assuming you meant ${0,1}$ and not the open interval $(0,1)$ with my edit. Note that we like descriptive titles here -- the more specific the better! Welcome to mathSE
$endgroup$
– 6005
Jan 11 at 17:52
add a comment |
$begingroup$
I am playing around with subgroups and I was wondering if the group ${0,1}$ is a subgroup of $mathbb{R}$ with respect to multiplication. It seems to fit the criteria but the inverse property is making me worry because $0$ is its own inverse. I vaguely remember a proof that if an element was its own inverse than it must be the identity element but $1$ would need to be the identity here. Thanks in advance!
abstract-algebra group-theory finite-groups
edited Jan 11 at 17:52
6005
36.2k751125
asked Jan 11 at 17:13
Ed W.Ed W.
112
$endgroup$
I am playing around with subgroups and I was wondering if the group ${0,1}$ is a subgroup of $mathbb{R}$ with respect to multiplication. It seems to fit the criteria but the inverse property is making me worry because $0$ is its own inverse. I vaguely remember a proof that if an element was its own inverse than it must be the identity element but $1$ would need to be the identity here. Thanks in advance!
abstract-algebra group-theory finite-groups
abstract-algebra group-theory finite-groups
edited Jan 11 at 17:52
6005
36.2k751125
asked Jan 11 at 17:13
Ed W.Ed W.
112
edited Jan 11 at 17:52
6005
36.2k751125
asked Jan 11 at 17:13
Ed W.Ed W.
112
edited Jan 11 at 17:52
6005
36.2k751125
edited Jan 11 at 17:52
6005
36.2k751125
edited Jan 11 at 17:52
6005
36.2k751125
36.2k751125
asked Jan 11 at 17:13
Ed W.Ed W.
112
asked Jan 11 at 17:13
Ed W.Ed W.
112
asked Jan 11 at 17:13
Ed W.Ed W.
112
112
2
$begingroup$
Assuming you mean the set ${0,1}subseteq mathbb{R}$ and $(mathbb{R},cdot)$ were a group under multiplication (which is not !). If $0$ were to have a multiplicative inverse then $exists; bin{0,1};:;0cdot b=1$. Is this possible? And if $a=a^{-1}$ then $a$ need not be identity. Consider the Klein-$4$ group ${1,-1,i,-i}$.
$endgroup$
– Yadati Kiran
Jan 11 at 17:16
$begingroup$
I see, I definitely need to do some more brushing up on this, thank you!
$endgroup$
– Ed W.
Jan 11 at 17:26
$begingroup$
I am assuming you meant ${0,1}$ and not the open interval $(0,1)$ with my edit. Note that we like descriptive titles here -- the more specific the better! Welcome to mathSE
$endgroup$
– 6005
Jan 11 at 17:52
add a comment |
2
$begingroup$
Assuming you mean the set ${0,1}subseteq mathbb{R}$ and $(mathbb{R},cdot)$ were a group under multiplication (which is not !). If $0$ were to have a multiplicative inverse then $exists; bin{0,1};:;0cdot b=1$. Is this possible? And if $a=a^{-1}$ then $a$ need not be identity. Consider the Klein-$4$ group ${1,-1,i,-i}$.
$endgroup$
– Yadati Kiran
Jan 11 at 17:16
$begingroup$
I see, I definitely need to do some more brushing up on this, thank you!
$endgroup$
– Ed W.
Jan 11 at 17:26
$begingroup$
I am assuming you meant ${0,1}$ and not the open interval $(0,1)$ with my edit. Note that we like descriptive titles here -- the more specific the better! Welcome to mathSE
$endgroup$
– 6005
Jan 11 at 17:52
2
2
$begingroup$
Assuming you mean the set ${0,1}subseteq mathbb{R}$ and $(mathbb{R},cdot)$ were a group under multiplication (which is not !). If $0$ were to have a multiplicative inverse then $exists; bin{0,1};:;0cdot b=1$. Is this possible? And if $a=a^{-1}$ then $a$ need not be identity. Consider the Klein-$4$ group ${1,-1,i,-i}$.
$endgroup$
– Yadati Kiran
Jan 11 at 17:16
$begingroup$
Assuming you mean the set ${0,1}subseteq mathbb{R}$ and $(mathbb{R},cdot)$ were a group under multiplication (which is not !). If $0$ were to have a multiplicative inverse then $exists; bin{0,1};:;0cdot b=1$. Is this possible? And if $a=a^{-1}$ then $a$ need not be identity. Consider the Klein-$4$ group ${1,-1,i,-i}$.
$endgroup$
– Yadati Kiran
Jan 11 at 17:16
$begingroup$
I see, I definitely need to do some more brushing up on this, thank you!
$endgroup$
– Ed W.
Jan 11 at 17:26
$begingroup$
I see, I definitely need to do some more brushing up on this, thank you!
$endgroup$
– Ed W.
Jan 11 at 17:26
$begingroup$
I am assuming you meant ${0,1}$ and not the open interval $(0,1)$ with my edit. Note that we like descriptive titles here -- the more specific the better! Welcome to mathSE
$endgroup$
– 6005
Jan 11 at 17:52
$begingroup$
I am assuming you meant ${0,1}$ and not the open interval $(0,1)$ with my edit. Note that we like descriptive titles here -- the more specific the better! Welcome to mathSE
$endgroup$
– 6005
Jan 11 at 17:52
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let's begin from the fact that $mathbb{R}$ is not a group with respect to multiplication because $0$ has no inverse at all. So of course it can't have any subgroups. Now, if you look at $mathbb{R}setminus{0}$ then it is a group with respect to multiplication. And no, it is not true that if an element is its own inverse then it is the identity element. For example in the group I mentioned $(-1)$ is its own inverse.
answered Jan 11 at 17:16
MarkMark
7,108417
$endgroup$
$begingroup$
Ok so the first assumption was wrong, thank you!
$endgroup$
– Ed W.
Jan 11 at 17:19
add a comment |
$begingroup$
To reiterate Mark's good answer, ${0,1}$ is not a subgroup of $mathbb{R}$ for two reasons:
$mathbb{R}$ is not a group under multiplication. In order for $A$ to be a subgroup of $B$, $B$ has to be a group.
($mathbb{R}$ is only a group under addition. Also, $mathbb{R} setminus {0}$ is a group under multiplication. But $mathbb{R}$ under multiplication isn't.)
${0,1}$ is not a group under multiplication itself.
It seems to fit the criteria but the inverse property is making me worry because 0 is its own inverse. I vaguely remember a proof that if an element was its own inverse than it must be the identity element but 1 would need to be the identity here.
Partly correct. $0$ isn't its own inverse -- $1$ is the only identity element in ${0,1}$ with addition, and $0$ isn't an identity, so $0$ wouldn't be called its own inverse.
But what is true -- and seems close to the reasoning you are getting at -- is that if $a cdot a = a$ in a group, then $a$ is the identity. This is because you can multiply on both sides by $a^{-1}$ and then you get $a = e$ (where $e$ denotes the identity).
Here, $0 cdot 0 = 0$, so $0$ would have to be the identity. But we know that doesn't work, since $1$ is already the identity, and $0 cdot 1 = 0$, not $1$.
answered Jan 11 at 17:32
60056005
36.2k751125
$endgroup$
add a comment |
$begingroup$
Hint: The Cayley table of ${0, 1}$ under multiplication is
$$begin{array}{c| c c}
times & 0 & 1\
hline
0 & 0 & 0 \
1 & 0 & 1.
end{array}$$
What do you know about Latin squares?
answered Jan 12 at 14:35
ShaunShaun
9,083113683
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's begin from the fact that $mathbb{R}$ is not a group with respect to multiplication because $0$ has no inverse at all. So of course it can't have any subgroups. Now, if you look at $mathbb{R}setminus{0}$ then it is a group with respect to multiplication. And no, it is not true that if an element is its own inverse then it is the identity element. For example in the group I mentioned $(-1)$ is its own inverse.
answered Jan 11 at 17:16
MarkMark
7,108417
$endgroup$
$begingroup$
Ok so the first assumption was wrong, thank you!
$endgroup$
– Ed W.
Jan 11 at 17:19
add a comment |
$begingroup$
Let's begin from the fact that $mathbb{R}$ is not a group with respect to multiplication because $0$ has no inverse at all. So of course it can't have any subgroups. Now, if you look at $mathbb{R}setminus{0}$ then it is a group with respect to multiplication. And no, it is not true that if an element is its own inverse then it is the identity element. For example in the group I mentioned $(-1)$ is its own inverse.
answered Jan 11 at 17:16
MarkMark
7,108417
$endgroup$
$begingroup$
Ok so the first assumption was wrong, thank you!
$endgroup$
– Ed W.
Jan 11 at 17:19
add a comment |
$begingroup$
Let's begin from the fact that $mathbb{R}$ is not a group with respect to multiplication because $0$ has no inverse at all. So of course it can't have any subgroups. Now, if you look at $mathbb{R}setminus{0}$ then it is a group with respect to multiplication. And no, it is not true that if an element is its own inverse then it is the identity element. For example in the group I mentioned $(-1)$ is its own inverse.
answered Jan 11 at 17:16
MarkMark
7,108417
$endgroup$
Let's begin from the fact that $mathbb{R}$ is not a group with respect to multiplication because $0$ has no inverse at all. So of course it can't have any subgroups. Now, if you look at $mathbb{R}setminus{0}$ then it is a group with respect to multiplication. And no, it is not true that if an element is its own inverse then it is the identity element. For example in the group I mentioned $(-1)$ is its own inverse.
answered Jan 11 at 17:16
MarkMark
7,108417
answered Jan 11 at 17:16
MarkMark
7,108417
answered Jan 11 at 17:16
MarkMark
7,108417
answered Jan 11 at 17:16
MarkMark
7,108417
7,108417
$begingroup$
Ok so the first assumption was wrong, thank you!
$endgroup$
– Ed W.
Jan 11 at 17:19
add a comment |
$begingroup$
Ok so the first assumption was wrong, thank you!
$endgroup$
– Ed W.
Jan 11 at 17:19
$begingroup$
Ok so the first assumption was wrong, thank you!
$endgroup$
– Ed W.
Jan 11 at 17:19
$begingroup$
Ok so the first assumption was wrong, thank you!
$endgroup$
– Ed W.
Jan 11 at 17:19
add a comment |
$begingroup$
To reiterate Mark's good answer, ${0,1}$ is not a subgroup of $mathbb{R}$ for two reasons:
$mathbb{R}$ is not a group under multiplication. In order for $A$ to be a subgroup of $B$, $B$ has to be a group.
($mathbb{R}$ is only a group under addition. Also, $mathbb{R} setminus {0}$ is a group under multiplication. But $mathbb{R}$ under multiplication isn't.)
${0,1}$ is not a group under multiplication itself.
It seems to fit the criteria but the inverse property is making me worry because 0 is its own inverse. I vaguely remember a proof that if an element was its own inverse than it must be the identity element but 1 would need to be the identity here.
Partly correct. $0$ isn't its own inverse -- $1$ is the only identity element in ${0,1}$ with addition, and $0$ isn't an identity, so $0$ wouldn't be called its own inverse.
But what is true -- and seems close to the reasoning you are getting at -- is that if $a cdot a = a$ in a group, then $a$ is the identity. This is because you can multiply on both sides by $a^{-1}$ and then you get $a = e$ (where $e$ denotes the identity).
Here, $0 cdot 0 = 0$, so $0$ would have to be the identity. But we know that doesn't work, since $1$ is already the identity, and $0 cdot 1 = 0$, not $1$.
answered Jan 11 at 17:32
60056005
36.2k751125
$endgroup$
add a comment |
$begingroup$
To reiterate Mark's good answer, ${0,1}$ is not a subgroup of $mathbb{R}$ for two reasons:
$mathbb{R}$ is not a group under multiplication. In order for $A$ to be a subgroup of $B$, $B$ has to be a group.
($mathbb{R}$ is only a group under addition. Also, $mathbb{R} setminus {0}$ is a group under multiplication. But $mathbb{R}$ under multiplication isn't.)
${0,1}$ is not a group under multiplication itself.
It seems to fit the criteria but the inverse property is making me worry because 0 is its own inverse. I vaguely remember a proof that if an element was its own inverse than it must be the identity element but 1 would need to be the identity here.
Partly correct. $0$ isn't its own inverse -- $1$ is the only identity element in ${0,1}$ with addition, and $0$ isn't an identity, so $0$ wouldn't be called its own inverse.
But what is true -- and seems close to the reasoning you are getting at -- is that if $a cdot a = a$ in a group, then $a$ is the identity. This is because you can multiply on both sides by $a^{-1}$ and then you get $a = e$ (where $e$ denotes the identity).
Here, $0 cdot 0 = 0$, so $0$ would have to be the identity. But we know that doesn't work, since $1$ is already the identity, and $0 cdot 1 = 0$, not $1$.
answered Jan 11 at 17:32
60056005
36.2k751125
$endgroup$
add a comment |
$begingroup$
To reiterate Mark's good answer, ${0,1}$ is not a subgroup of $mathbb{R}$ for two reasons:
$mathbb{R}$ is not a group under multiplication. In order for $A$ to be a subgroup of $B$, $B$ has to be a group.
($mathbb{R}$ is only a group under addition. Also, $mathbb{R} setminus {0}$ is a group under multiplication. But $mathbb{R}$ under multiplication isn't.)
${0,1}$ is not a group under multiplication itself.
It seems to fit the criteria but the inverse property is making me worry because 0 is its own inverse. I vaguely remember a proof that if an element was its own inverse than it must be the identity element but 1 would need to be the identity here.
Partly correct. $0$ isn't its own inverse -- $1$ is the only identity element in ${0,1}$ with addition, and $0$ isn't an identity, so $0$ wouldn't be called its own inverse.
But what is true -- and seems close to the reasoning you are getting at -- is that if $a cdot a = a$ in a group, then $a$ is the identity. This is because you can multiply on both sides by $a^{-1}$ and then you get $a = e$ (where $e$ denotes the identity).
Here, $0 cdot 0 = 0$, so $0$ would have to be the identity. But we know that doesn't work, since $1$ is already the identity, and $0 cdot 1 = 0$, not $1$.
answered Jan 11 at 17:32
60056005
36.2k751125
$endgroup$
To reiterate Mark's good answer, ${0,1}$ is not a subgroup of $mathbb{R}$ for two reasons:
$mathbb{R}$ is not a group under multiplication. In order for $A$ to be a subgroup of $B$, $B$ has to be a group.
($mathbb{R}$ is only a group under addition. Also, $mathbb{R} setminus {0}$ is a group under multiplication. But $mathbb{R}$ under multiplication isn't.)
${0,1}$ is not a group under multiplication itself.
It seems to fit the criteria but the inverse property is making me worry because 0 is its own inverse. I vaguely remember a proof that if an element was its own inverse than it must be the identity element but 1 would need to be the identity here.
Partly correct. $0$ isn't its own inverse -- $1$ is the only identity element in ${0,1}$ with addition, and $0$ isn't an identity, so $0$ wouldn't be called its own inverse.
But what is true -- and seems close to the reasoning you are getting at -- is that if $a cdot a = a$ in a group, then $a$ is the identity. This is because you can multiply on both sides by $a^{-1}$ and then you get $a = e$ (where $e$ denotes the identity).
Here, $0 cdot 0 = 0$, so $0$ would have to be the identity. But we know that doesn't work, since $1$ is already the identity, and $0 cdot 1 = 0$, not $1$.
answered Jan 11 at 17:32
60056005
36.2k751125
answered Jan 11 at 17:32
60056005
36.2k751125
answered Jan 11 at 17:32
60056005
36.2k751125
answered Jan 11 at 17:32
60056005
36.2k751125
36.2k751125
add a comment |
add a comment |
$begingroup$
Hint: The Cayley table of ${0, 1}$ under multiplication is
$$begin{array}{c| c c}
times & 0 & 1\
hline
0 & 0 & 0 \
1 & 0 & 1.
end{array}$$
What do you know about Latin squares?
answered Jan 12 at 14:35
ShaunShaun
9,083113683
$endgroup$
add a comment |
$begingroup$
Hint: The Cayley table of ${0, 1}$ under multiplication is
$$begin{array}{c| c c}
times & 0 & 1\
hline
0 & 0 & 0 \
1 & 0 & 1.
end{array}$$
What do you know about Latin squares?
answered Jan 12 at 14:35
ShaunShaun
9,083113683
$endgroup$
add a comment |
$begingroup$
Hint: The Cayley table of ${0, 1}$ under multiplication is
$$begin{array}{c| c c}
times & 0 & 1\
hline
0 & 0 & 0 \
1 & 0 & 1.
end{array}$$
What do you know about Latin squares?
answered Jan 12 at 14:35
ShaunShaun
9,083113683
$endgroup$
Hint: The Cayley table of ${0, 1}$ under multiplication is
$$begin{array}{c| c c}
times & 0 & 1\
hline
0 & 0 & 0 \
1 & 0 & 1.
end{array}$$
What do you know about Latin squares?
answered Jan 12 at 14:35
ShaunShaun
9,083113683
edited Jan 15 at 10:30
answered Jan 12 at 14:35
ShaunShaun
9,083113683
answered Jan 12 at 14:35
ShaunShaun
9,083113683
answered Jan 12 at 14:35
ShaunShaun
9,083113683
9,083113683
add a comment |
add a comment |
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$begingroup$
Assuming you mean the set ${0,1}subseteq mathbb{R}$ and $(mathbb{R},cdot)$ were a group under multiplication (which is not !). If $0$ were to have a multiplicative inverse then $exists; bin{0,1};:;0cdot b=1$. Is this possible? And if $a=a^{-1}$ then $a$ need not be identity. Consider the Klein-$4$ group ${1,-1,i,-i}$.
$endgroup$
– Yadati Kiran
Jan 11 at 17:16
$begingroup$
I see, I definitely need to do some more brushing up on this, thank you!
$endgroup$
– Ed W.
Jan 11 at 17:26
$begingroup$
I am assuming you meant ${0,1}$ and not the open interval $(0,1)$ with my edit. Note that we like descriptive titles here -- the more specific the better! Welcome to mathSE
$endgroup$
– 6005
Jan 11 at 17:52