In a group G, show that the equation ax=b has only one solution for each a,b in GHow to show that the group of nonzero real numbers under multiplication is not isomorphic to the group of real numbers under addition?Show that the set $r^-1 : rin R $ contains exactly one element out of each right coset of $H$.If $G$ is a group, show that $x^2ax=a^-1$ has a solution if and only if $a$ is a cube in $G$A finite abelian group $G$ has invariant factors $(m_1, m_2, …, m_k)$. Show that $G$ has an element of order $s$ if and only if $s$ divides $m_k$.Suppose equation $x^12 = 1$ has $14$ solution in some group. Show that this group is not cyclic.How do I show that there exists only one group of order 2 up to isomorphism?Show that a nonabelian group G of order $pq$, with $p$ and $q$ two different prime numbers, is solvable but not nilpotent.Prove, that equation xaxba=xbc has only one solution, for firmly selected a,b,c, from group.Use the Euclidean algorithm to show 74 and 105 are relatively prime. Be sure to identify at least one solution to the equation 105s+74t=1Why the equation in Burnside's lemma has only one solution
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In a group G, show that the equation ax=b has only one solution for each a,b in G
How to show that the group of nonzero real numbers under multiplication is not isomorphic to the group of real numbers under addition?Show that the set $r^-1 : rin R $ contains exactly one element out of each right coset of $H$.If $G$ is a group, show that $x^2ax=a^-1$ has a solution if and only if $a$ is a cube in $G$A finite abelian group $G$ has invariant factors $(m_1, m_2, …, m_k)$. Show that $G$ has an element of order $s$ if and only if $s$ divides $m_k$.Suppose equation $x^12 = 1$ has $14$ solution in some group. Show that this group is not cyclic.How do I show that there exists only one group of order 2 up to isomorphism?Show that a nonabelian group G of order $pq$, with $p$ and $q$ two different prime numbers, is solvable but not nilpotent.Prove, that equation xaxba=xbc has only one solution, for firmly selected a,b,c, from group.Use the Euclidean algorithm to show 74 and 105 are relatively prime. Be sure to identify at least one solution to the equation 105s+74t=1Why the equation in Burnside's lemma has only one solution
$begingroup$
I see there are two things to show. First, that there exists a solution, and second, that the solution is the only one, but I'm not sure how to do it.
abstract-algebra group-theory
asked 19 hours ago
NickNick
294
$endgroup$
add a comment |
$begingroup$
I see there are two things to show. First, that there exists a solution, and second, that the solution is the only one, but I'm not sure how to do it.
abstract-algebra group-theory
asked 19 hours ago
NickNick
294
$endgroup$
2
$begingroup$
The context is missing in your question. If your definition of a group includes this as one of the axioms, then there is nothing to show. You are using the "usual" axioms of a group, correct?
$endgroup$
– Somos
19 hours ago
add a comment |
$begingroup$
I see there are two things to show. First, that there exists a solution, and second, that the solution is the only one, but I'm not sure how to do it.
abstract-algebra group-theory
asked 19 hours ago
NickNick
294
$endgroup$
I see there are two things to show. First, that there exists a solution, and second, that the solution is the only one, but I'm not sure how to do it.
abstract-algebra group-theory
abstract-algebra group-theory
asked 19 hours ago
NickNick
294
asked 19 hours ago
NickNick
294
asked 19 hours ago
NickNick
294
asked 19 hours ago
NickNick
294
asked 19 hours ago
NickNick
294
294
2
$begingroup$
The context is missing in your question. If your definition of a group includes this as one of the axioms, then there is nothing to show. You are using the "usual" axioms of a group, correct?
$endgroup$
– Somos
19 hours ago
add a comment |
2
$begingroup$
The context is missing in your question. If your definition of a group includes this as one of the axioms, then there is nothing to show. You are using the "usual" axioms of a group, correct?
$endgroup$
– Somos
19 hours ago
2
2
$begingroup$
The context is missing in your question. If your definition of a group includes this as one of the axioms, then there is nothing to show. You are using the "usual" axioms of a group, correct?
$endgroup$
– Somos
19 hours ago
$begingroup$
The context is missing in your question. If your definition of a group includes this as one of the axioms, then there is nothing to show. You are using the "usual" axioms of a group, correct?
$endgroup$
– Somos
19 hours ago
add a comment |
1 Answer
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$begingroup$
$ax = b implies a^-1ax = a^-1b implies x = a^-1b$. The solution is unique since the inverse is unique.
answered 19 hours ago
Kaixin WangKaixin Wang
1
New contributor
Kaixin Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
Hmm - I feel that for full marks you'd have to prove that the inverse is unique! I mean, all you have done is reduce it to a specific instance of the problem: $ax=1$.
$endgroup$
– user1729
13 hours ago
add a comment |
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$begingroup$
$ax = b implies a^-1ax = a^-1b implies x = a^-1b$. The solution is unique since the inverse is unique.
answered 19 hours ago
Kaixin WangKaixin Wang
1
New contributor
Kaixin Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
Hmm - I feel that for full marks you'd have to prove that the inverse is unique! I mean, all you have done is reduce it to a specific instance of the problem: $ax=1$.
$endgroup$
– user1729
13 hours ago
add a comment |
$begingroup$
$ax = b implies a^-1ax = a^-1b implies x = a^-1b$. The solution is unique since the inverse is unique.
answered 19 hours ago
Kaixin WangKaixin Wang
1
New contributor
Kaixin Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
Hmm - I feel that for full marks you'd have to prove that the inverse is unique! I mean, all you have done is reduce it to a specific instance of the problem: $ax=1$.
$endgroup$
– user1729
13 hours ago
add a comment |
$begingroup$
$ax = b implies a^-1ax = a^-1b implies x = a^-1b$. The solution is unique since the inverse is unique.
answered 19 hours ago
Kaixin WangKaixin Wang
1
New contributor
Kaixin Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$ax = b implies a^-1ax = a^-1b implies x = a^-1b$. The solution is unique since the inverse is unique.
answered 19 hours ago
Kaixin WangKaixin Wang
1
New contributor
Kaixin Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 19 hours ago
Kaixin WangKaixin Wang
1
New contributor
Kaixin Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 19 hours ago
Kaixin WangKaixin Wang
1
answered 19 hours ago
Kaixin WangKaixin Wang
1
1
New contributor
Kaixin Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Kaixin Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Kaixin Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
Hmm - I feel that for full marks you'd have to prove that the inverse is unique! I mean, all you have done is reduce it to a specific instance of the problem: $ax=1$.
$endgroup$
– user1729
13 hours ago
add a comment |
2
$begingroup$
Hmm - I feel that for full marks you'd have to prove that the inverse is unique! I mean, all you have done is reduce it to a specific instance of the problem: $ax=1$.
$endgroup$
– user1729
13 hours ago
2
2
$begingroup$
Hmm - I feel that for full marks you'd have to prove that the inverse is unique! I mean, all you have done is reduce it to a specific instance of the problem: $ax=1$.
$endgroup$
– user1729
13 hours ago
$begingroup$
Hmm - I feel that for full marks you'd have to prove that the inverse is unique! I mean, all you have done is reduce it to a specific instance of the problem: $ax=1$.
$endgroup$
– user1729
13 hours ago
add a comment |
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$begingroup$
The context is missing in your question. If your definition of a group includes this as one of the axioms, then there is nothing to show. You are using the "usual" axioms of a group, correct?
$endgroup$
– Somos
19 hours ago