group iff $G$ is a set, $*$ associative binary operation satisfying left identity, left inverseSets, groups and generatorsProve a set with this associative operation is a groupFor an associative binary operation with identity, the set of invertible elements forms a groupDoes the set operation define a binary operation on G?A set with a binary operation and an interesting property defines a groupFind a binary operation on a set of $5$ elements satisfying certain conditions and which does not define a groupNon-associative commutative binary operationHow to prove identity element in a set that is finite, associative, and left-right cancellable?Non-associative, non-commutative binary operation with a identity elementIn a group, the identity of the binary operation must be unique?Left identity and left inverse on set makes it the group
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group iff $G$ is a set, $*$ associative binary operation satisfying left identity, left inverse
Sets, groups and generatorsProve a set with this associative operation is a groupFor an associative binary operation with identity, the set of invertible elements forms a groupDoes the set operation define a binary operation on G?A set with a binary operation and an interesting property defines a groupFind a binary operation on a set of $5$ elements satisfying certain conditions and which does not define a groupNon-associative commutative binary operationHow to prove identity element in a set that is finite, associative, and left-right cancellable?Non-associative, non-commutative binary operation with a identity elementIn a group, the identity of the binary operation must be unique?Left identity and left inverse on set makes it the group
$begingroup$
How would I go about showing that the pair $(G, *)$ is a group if and only if $G$ is a set and $*$ is an associative binary operation on $G$ such that:
- (Left Identity) There exists an element $e in G$ such that $e * g = g$ for all $g in G$.
- (Left Inverses) For each $g in G$, there exists an element $g^-1 in G$ such that $g^-1 * g = e$.
Any help would be greatly appreciated, thanks!
abstract-algebra group-theory
$endgroup$
add a comment |
$begingroup$
How would I go about showing that the pair $(G, *)$ is a group if and only if $G$ is a set and $*$ is an associative binary operation on $G$ such that:
- (Left Identity) There exists an element $e in G$ such that $e * g = g$ for all $g in G$.
- (Left Inverses) For each $g in G$, there exists an element $g^-1 in G$ such that $g^-1 * g = e$.
Any help would be greatly appreciated, thanks!
abstract-algebra group-theory
$endgroup$
add a comment |
$begingroup$
How would I go about showing that the pair $(G, *)$ is a group if and only if $G$ is a set and $*$ is an associative binary operation on $G$ such that:
- (Left Identity) There exists an element $e in G$ such that $e * g = g$ for all $g in G$.
- (Left Inverses) For each $g in G$, there exists an element $g^-1 in G$ such that $g^-1 * g = e$.
Any help would be greatly appreciated, thanks!
abstract-algebra group-theory
$endgroup$
How would I go about showing that the pair $(G, *)$ is a group if and only if $G$ is a set and $*$ is an associative binary operation on $G$ such that:
- (Left Identity) There exists an element $e in G$ such that $e * g = g$ for all $g in G$.
- (Left Inverses) For each $g in G$, there exists an element $g^-1 in G$ such that $g^-1 * g = e$.
Any help would be greatly appreciated, thanks!
abstract-algebra group-theory
abstract-algebra group-theory
asked Dec 12 '14 at 2:46
user200570
add a comment |
add a comment |
1 Answer
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$begingroup$
If $G$ is a group, conditions $(1)$ and $(2)$ are directly implied by the axioms. So suppose now that $(1)$ and $(2)$ are true. We need to show existence of right inverses and that the identity also works from the right.
Identity
Note that $$g^-1 * g = e = e * e = g^-1 * g * e.$$ By $(2)$ there is $h in G$ such that $h * g^-1 = e$. Thus $$h * g^-1 * g = h * g^-1 * g * e text Longrightarrow text e* g = e* g * e = e text Longrightarrowtext g = g * e.$$
Inverses
We have $$g^-1 * g * g^-1 = (g^-1 * g) * g^-1 = e * g^-1 = g^-1 text Longrightarrow text h * g^-1 * g * g^-1 = h * g^-1$$$$Longrightarrowtext e * (g * g^-1) = etext Longrightarrow text g * g^-1 = e.$$
answered Dec 12 '14 at 2:54
user149792user149792
4,15342470
$endgroup$
add a comment |
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1 Answer
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$begingroup$
If $G$ is a group, conditions $(1)$ and $(2)$ are directly implied by the axioms. So suppose now that $(1)$ and $(2)$ are true. We need to show existence of right inverses and that the identity also works from the right.
Identity
Note that $$g^-1 * g = e = e * e = g^-1 * g * e.$$ By $(2)$ there is $h in G$ such that $h * g^-1 = e$. Thus $$h * g^-1 * g = h * g^-1 * g * e text Longrightarrow text e* g = e* g * e = e text Longrightarrowtext g = g * e.$$
Inverses
We have $$g^-1 * g * g^-1 = (g^-1 * g) * g^-1 = e * g^-1 = g^-1 text Longrightarrow text h * g^-1 * g * g^-1 = h * g^-1$$$$Longrightarrowtext e * (g * g^-1) = etext Longrightarrow text g * g^-1 = e.$$
answered Dec 12 '14 at 2:54
user149792user149792
4,15342470
$endgroup$
add a comment |
$begingroup$
If $G$ is a group, conditions $(1)$ and $(2)$ are directly implied by the axioms. So suppose now that $(1)$ and $(2)$ are true. We need to show existence of right inverses and that the identity also works from the right.
Identity
Note that $$g^-1 * g = e = e * e = g^-1 * g * e.$$ By $(2)$ there is $h in G$ such that $h * g^-1 = e$. Thus $$h * g^-1 * g = h * g^-1 * g * e text Longrightarrow text e* g = e* g * e = e text Longrightarrowtext g = g * e.$$
Inverses
We have $$g^-1 * g * g^-1 = (g^-1 * g) * g^-1 = e * g^-1 = g^-1 text Longrightarrow text h * g^-1 * g * g^-1 = h * g^-1$$$$Longrightarrowtext e * (g * g^-1) = etext Longrightarrow text g * g^-1 = e.$$
answered Dec 12 '14 at 2:54
user149792user149792
4,15342470
$endgroup$
add a comment |
$begingroup$
If $G$ is a group, conditions $(1)$ and $(2)$ are directly implied by the axioms. So suppose now that $(1)$ and $(2)$ are true. We need to show existence of right inverses and that the identity also works from the right.
Identity
Note that $$g^-1 * g = e = e * e = g^-1 * g * e.$$ By $(2)$ there is $h in G$ such that $h * g^-1 = e$. Thus $$h * g^-1 * g = h * g^-1 * g * e text Longrightarrow text e* g = e* g * e = e text Longrightarrowtext g = g * e.$$
Inverses
We have $$g^-1 * g * g^-1 = (g^-1 * g) * g^-1 = e * g^-1 = g^-1 text Longrightarrow text h * g^-1 * g * g^-1 = h * g^-1$$$$Longrightarrowtext e * (g * g^-1) = etext Longrightarrow text g * g^-1 = e.$$
answered Dec 12 '14 at 2:54
user149792user149792
4,15342470
$endgroup$
If $G$ is a group, conditions $(1)$ and $(2)$ are directly implied by the axioms. So suppose now that $(1)$ and $(2)$ are true. We need to show existence of right inverses and that the identity also works from the right.
Identity
Note that $$g^-1 * g = e = e * e = g^-1 * g * e.$$ By $(2)$ there is $h in G$ such that $h * g^-1 = e$. Thus $$h * g^-1 * g = h * g^-1 * g * e text Longrightarrow text e* g = e* g * e = e text Longrightarrowtext g = g * e.$$
Inverses
We have $$g^-1 * g * g^-1 = (g^-1 * g) * g^-1 = e * g^-1 = g^-1 text Longrightarrow text h * g^-1 * g * g^-1 = h * g^-1$$$$Longrightarrowtext e * (g * g^-1) = etext Longrightarrow text g * g^-1 = e.$$
answered Dec 12 '14 at 2:54
user149792user149792
4,15342470
edited Dec 12 '14 at 3:07
answered Dec 12 '14 at 2:54
user149792user149792
4,15342470
answered Dec 12 '14 at 2:54
user149792user149792
4,15342470
answered Dec 12 '14 at 2:54
user149792user149792
4,15342470
4,15342470
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