Linked Questions

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3answers
13k views

Homework - Prove that a given set is a group [duplicate]

I have a homework question and I don't know how to approach this exercise. The exercise is the following: Let's suppose $G$ is a set with binary function * defined for its members, which is: ...
3
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1answer
1k views

Finding the Right Inverses and Right Identity of a Group [duplicate]

Possible Duplicate: Right identity and Right inverse implies a group Let $(G,*)$ be a binary structure that has the following properties: 1) The binary operation $*$ is associative. 2) There ...
0
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0answers
57 views

Is $G$ a group under those conditions? [duplicate]

Let $G$ be a set and $*:G\times G \rightarrow G$ an operation with: (i) $(G,*)$ is associative (ii) There is a $f \in G$ with $f*a=a$ for all $a \in G$ (iii) For every $a \in G$ exists a $b \in G$ ...
2
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0answers
37 views

Is the existence of a right identity and right inverses a sufficient condition for $(G,*)$ to be a group? [duplicate]

I have come across an exercise that I cannot solve and honestly, to me, it doesn’t even seem true. It goes like this: Let $G$ be a non empty set and $*$ an associative operation on $G$ such that: $...
66
votes
10answers
9k views

Why not include as a requirement that all functions must be continuous to be differentiable?

Theorem: Suppose that $f : A \to \mathbb{R}$ where $A \subseteq \mathbb{R}$. If $f$ is differentiable at $x \in A$, then $f$ is continuous at $x$. This theorem is equivalent (by the contrapositive) ...
11
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5answers
2k views

Are there algebraic structures with more than one neutral element and/or more than one inverse element?

I was reading a book on groups, it points out about the uniqueness of the neutral element and the inverse element. I got curious, are there algebraic structures with more than one neutral element and/...
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3answers
6k views

Any Set with Associativity, Left Identity, Left Inverse is a Group

Related Link: Right identity and Right inverse implies a group Reference: Fraleigh p. 49 Question 4.38 in A First Course in Abstract Algebra I will present my proof (distinct from those in the ...
2
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2answers
4k views

To prove in a Group Left identity and left inverse implies right identity and right inverse

Let $G$ be a nonempty set closed under an associative product, which in addition satisfies : A. There exists an $e$ in $G$ such that $a \cdot e=a$ for all $a \in G$. B. Given $a \in G$, ...
4
votes
2answers
2k views

Proving that if $ab=e$ then $ba=e$

Suppose that instead of the property $ab=ba=e$ a group G has the condition that for every element $a$ there exists an element $b$, such that $ab=e$. Prove that $ba=e$. Is the following a valid proof? ...
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0answers
1k views

Let G be a group and suppose $a*b*c=e$ for some $a,b,c \in G$. Then $b*c*a=e$ as well.

Just as the title says: Proposition. Let G be a group and suppose $a*b*c=e$ for some $a,b,c \in G$. Then $b*c*a=e$ as well. I'm looking for a proof review, please. Here is my proof: Proof. If $a*b*...
4
votes
1answer
229 views

Proof of brauer's lemma, $eRe$ being a division ring.

On page 1 of this article, the author proves the following claim: Brauer's Lemma: Let $K$ be a minimal left ideal of a ring $R$, with $K^2 \not= 0$. Then $K=Re$ where $e^2=e \in R$, and $eRe$ is a ...
2
votes
1answer
176 views

Prove that (G,*) is a group.

G is a monoid and satisfies the right inverse law. Show that G is a group. I tried next: It is obviously sufficient to show that G satisfies left inverse law. (I use $a'$ notation for $a$ inverse.) We ...
2
votes
1answer
101 views

Do those 'groups' have a name?

Is there a name for 'groups' that only have a neutral element on the right and an inverse for each element on the right ? If there is, does that name also hold for a neutral elt on the right and an ...
1
vote
2answers
156 views

If a magma M is both a semigroup and a quasigroup, is it necessarily a group?

If a magma which has an identity element is also a semigroup and a quasigroup, it can be shown that this is indeed a group. I'm looking for a counter example: a magma which is a quasigroup (for every ...
2
votes
1answer
81 views

Is a semigroup with unique right identity and left inverse a group?

We know that a semigroup with a right identity and right inverse for all elements is a group (e.g. see here). Symmetrically, also a left identity together with a left inverse implies a group. We also ...

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