How to show that $⟨a,b | aba=bab⟩$ is not the trivial group?
$begingroup$
I want to show that G = $⟨a,b | aba=bab⟩$ is not the trivial group
I tried to find homomorphism $phi$ from $G$ to $mathbb Z$ which maps $a$ to $0$ and $b$ to $1$ (or $b$ to $0$ and $a$ to $1$) and if such a homomorphism exists ,
$phi(b)$ is non-trivial and thus b is non-trivial.
but I didn't found such a homomorphism.
I'm also tried to conclude it directly from the relation and I failed again
Thanks
abstract-algebra group-theory finite-groups
edited Dec 7 '18 at 13:56
Andrews
3831317
asked Dec 7 '18 at 13:11
Roee PartoushRoee Partoush
111
$endgroup$
add a comment |
$begingroup$
I want to show that G = $⟨a,b | aba=bab⟩$ is not the trivial group
I tried to find homomorphism $phi$ from $G$ to $mathbb Z$ which maps $a$ to $0$ and $b$ to $1$ (or $b$ to $0$ and $a$ to $1$) and if such a homomorphism exists ,
$phi(b)$ is non-trivial and thus b is non-trivial.
but I didn't found such a homomorphism.
I'm also tried to conclude it directly from the relation and I failed again
Thanks
abstract-algebra group-theory finite-groups
edited Dec 7 '18 at 13:56
Andrews
3831317
asked Dec 7 '18 at 13:11
Roee PartoushRoee Partoush
111
$endgroup$
1
$begingroup$
If you add the relation $ab=ba$ then you quickly deduce $a=b$, so there is a surjective map to $mathbb Z$.
$endgroup$
– lulu
Dec 7 '18 at 13:13
$begingroup$
You can define your homomorphism to ${mathbb Z}$ by mapping $a$ and $b$ to $1$.
$endgroup$
– Derek Holt
Dec 7 '18 at 15:12
add a comment |
$begingroup$
I want to show that G = $⟨a,b | aba=bab⟩$ is not the trivial group
I tried to find homomorphism $phi$ from $G$ to $mathbb Z$ which maps $a$ to $0$ and $b$ to $1$ (or $b$ to $0$ and $a$ to $1$) and if such a homomorphism exists ,
$phi(b)$ is non-trivial and thus b is non-trivial.
but I didn't found such a homomorphism.
I'm also tried to conclude it directly from the relation and I failed again
Thanks
abstract-algebra group-theory finite-groups
edited Dec 7 '18 at 13:56
Andrews
3831317
asked Dec 7 '18 at 13:11
Roee PartoushRoee Partoush
111
$endgroup$
I want to show that G = $⟨a,b | aba=bab⟩$ is not the trivial group
I tried to find homomorphism $phi$ from $G$ to $mathbb Z$ which maps $a$ to $0$ and $b$ to $1$ (or $b$ to $0$ and $a$ to $1$) and if such a homomorphism exists ,
$phi(b)$ is non-trivial and thus b is non-trivial.
but I didn't found such a homomorphism.
I'm also tried to conclude it directly from the relation and I failed again
Thanks
abstract-algebra group-theory finite-groups
abstract-algebra group-theory finite-groups
edited Dec 7 '18 at 13:56
Andrews
3831317
asked Dec 7 '18 at 13:11
Roee PartoushRoee Partoush
111
edited Dec 7 '18 at 13:56
Andrews
3831317
asked Dec 7 '18 at 13:11
Roee PartoushRoee Partoush
111
edited Dec 7 '18 at 13:56
Andrews
3831317
edited Dec 7 '18 at 13:56
Andrews
3831317
edited Dec 7 '18 at 13:56
Andrews
3831317
3831317
asked Dec 7 '18 at 13:11
Roee PartoushRoee Partoush
111
asked Dec 7 '18 at 13:11
Roee PartoushRoee Partoush
111
asked Dec 7 '18 at 13:11
Roee PartoushRoee Partoush
111
111
1
$begingroup$
If you add the relation $ab=ba$ then you quickly deduce $a=b$, so there is a surjective map to $mathbb Z$.
$endgroup$
– lulu
Dec 7 '18 at 13:13
$begingroup$
You can define your homomorphism to ${mathbb Z}$ by mapping $a$ and $b$ to $1$.
$endgroup$
– Derek Holt
Dec 7 '18 at 15:12
add a comment |
1
$begingroup$
If you add the relation $ab=ba$ then you quickly deduce $a=b$, so there is a surjective map to $mathbb Z$.
$endgroup$
– lulu
Dec 7 '18 at 13:13
$begingroup$
You can define your homomorphism to ${mathbb Z}$ by mapping $a$ and $b$ to $1$.
$endgroup$
– Derek Holt
Dec 7 '18 at 15:12
1
1
$begingroup$
If you add the relation $ab=ba$ then you quickly deduce $a=b$, so there is a surjective map to $mathbb Z$.
$endgroup$
– lulu
Dec 7 '18 at 13:13
$begingroup$
If you add the relation $ab=ba$ then you quickly deduce $a=b$, so there is a surjective map to $mathbb Z$.
$endgroup$
– lulu
Dec 7 '18 at 13:13
$begingroup$
You can define your homomorphism to ${mathbb Z}$ by mapping $a$ and $b$ to $1$.
$endgroup$
– Derek Holt
Dec 7 '18 at 15:12
$begingroup$
You can define your homomorphism to ${mathbb Z}$ by mapping $a$ and $b$ to $1$.
$endgroup$
– Derek Holt
Dec 7 '18 at 15:12
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that $mathbb Z$ satisfies this, with $a=b$.
Phrased differently, adding the relation $ab=ba$ we quickly deduce that $a=b$, so the new relation gives a surjective map to $mathbb Z$.
answered Dec 7 '18 at 13:14
lulululu
39.8k24778
$endgroup$
$begingroup$
Also $S_3$ with $a=(1 2)$ and $b=(2 3)$.
$endgroup$
– bof
Dec 7 '18 at 14:41
$begingroup$
@lulu OK, so we can "extend" $G$ with another relation and get a presentaion of $Z$ , can you please explain how this helps in my case ? Tnx
$endgroup$
– Roee Partoush
Dec 7 '18 at 15:38
$begingroup$
If a set surjects onto an infinite set, it is obviously infinite.
$endgroup$
– lulu
Dec 7 '18 at 19:50
add a comment |
$begingroup$
The deficiency of a finite presentation $<X|R>$ is defined to be $|X|-|R|$. If $deg(G)>0$ then group $G$ is of order infinite.
answered Dec 7 '18 at 15:46
yavaryavar
843
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that $mathbb Z$ satisfies this, with $a=b$.
Phrased differently, adding the relation $ab=ba$ we quickly deduce that $a=b$, so the new relation gives a surjective map to $mathbb Z$.
answered Dec 7 '18 at 13:14
lulululu
39.8k24778
$endgroup$
$begingroup$
Also $S_3$ with $a=(1 2)$ and $b=(2 3)$.
$endgroup$
– bof
Dec 7 '18 at 14:41
$begingroup$
@lulu OK, so we can "extend" $G$ with another relation and get a presentaion of $Z$ , can you please explain how this helps in my case ? Tnx
$endgroup$
– Roee Partoush
Dec 7 '18 at 15:38
$begingroup$
If a set surjects onto an infinite set, it is obviously infinite.
$endgroup$
– lulu
Dec 7 '18 at 19:50
add a comment |
$begingroup$
Note that $mathbb Z$ satisfies this, with $a=b$.
Phrased differently, adding the relation $ab=ba$ we quickly deduce that $a=b$, so the new relation gives a surjective map to $mathbb Z$.
answered Dec 7 '18 at 13:14
lulululu
39.8k24778
$endgroup$
$begingroup$
Also $S_3$ with $a=(1 2)$ and $b=(2 3)$.
$endgroup$
– bof
Dec 7 '18 at 14:41
$begingroup$
@lulu OK, so we can "extend" $G$ with another relation and get a presentaion of $Z$ , can you please explain how this helps in my case ? Tnx
$endgroup$
– Roee Partoush
Dec 7 '18 at 15:38
$begingroup$
If a set surjects onto an infinite set, it is obviously infinite.
$endgroup$
– lulu
Dec 7 '18 at 19:50
add a comment |
$begingroup$
Note that $mathbb Z$ satisfies this, with $a=b$.
Phrased differently, adding the relation $ab=ba$ we quickly deduce that $a=b$, so the new relation gives a surjective map to $mathbb Z$.
answered Dec 7 '18 at 13:14
lulululu
39.8k24778
$endgroup$
Note that $mathbb Z$ satisfies this, with $a=b$.
Phrased differently, adding the relation $ab=ba$ we quickly deduce that $a=b$, so the new relation gives a surjective map to $mathbb Z$.
answered Dec 7 '18 at 13:14
lulululu
39.8k24778
answered Dec 7 '18 at 13:14
lulululu
39.8k24778
answered Dec 7 '18 at 13:14
lulululu
39.8k24778
answered Dec 7 '18 at 13:14
lulululu
39.8k24778
39.8k24778
$begingroup$
Also $S_3$ with $a=(1 2)$ and $b=(2 3)$.
$endgroup$
– bof
Dec 7 '18 at 14:41
$begingroup$
@lulu OK, so we can "extend" $G$ with another relation and get a presentaion of $Z$ , can you please explain how this helps in my case ? Tnx
$endgroup$
– Roee Partoush
Dec 7 '18 at 15:38
$begingroup$
If a set surjects onto an infinite set, it is obviously infinite.
$endgroup$
– lulu
Dec 7 '18 at 19:50
add a comment |
$begingroup$
Also $S_3$ with $a=(1 2)$ and $b=(2 3)$.
$endgroup$
– bof
Dec 7 '18 at 14:41
$begingroup$
@lulu OK, so we can "extend" $G$ with another relation and get a presentaion of $Z$ , can you please explain how this helps in my case ? Tnx
$endgroup$
– Roee Partoush
Dec 7 '18 at 15:38
$begingroup$
If a set surjects onto an infinite set, it is obviously infinite.
$endgroup$
– lulu
Dec 7 '18 at 19:50
$begingroup$
Also $S_3$ with $a=(1 2)$ and $b=(2 3)$.
$endgroup$
– bof
Dec 7 '18 at 14:41
$begingroup$
Also $S_3$ with $a=(1 2)$ and $b=(2 3)$.
$endgroup$
– bof
Dec 7 '18 at 14:41
$begingroup$
@lulu OK, so we can "extend" $G$ with another relation and get a presentaion of $Z$ , can you please explain how this helps in my case ? Tnx
$endgroup$
– Roee Partoush
Dec 7 '18 at 15:38
$begingroup$
@lulu OK, so we can "extend" $G$ with another relation and get a presentaion of $Z$ , can you please explain how this helps in my case ? Tnx
$endgroup$
– Roee Partoush
Dec 7 '18 at 15:38
$begingroup$
If a set surjects onto an infinite set, it is obviously infinite.
$endgroup$
– lulu
Dec 7 '18 at 19:50
$begingroup$
If a set surjects onto an infinite set, it is obviously infinite.
$endgroup$
– lulu
Dec 7 '18 at 19:50
add a comment |
$begingroup$
The deficiency of a finite presentation $<X|R>$ is defined to be $|X|-|R|$. If $deg(G)>0$ then group $G$ is of order infinite.
answered Dec 7 '18 at 15:46
yavaryavar
843
$endgroup$
add a comment |
$begingroup$
The deficiency of a finite presentation $<X|R>$ is defined to be $|X|-|R|$. If $deg(G)>0$ then group $G$ is of order infinite.
answered Dec 7 '18 at 15:46
yavaryavar
843
$endgroup$
add a comment |
$begingroup$
The deficiency of a finite presentation $<X|R>$ is defined to be $|X|-|R|$. If $deg(G)>0$ then group $G$ is of order infinite.
answered Dec 7 '18 at 15:46
yavaryavar
843
$endgroup$
The deficiency of a finite presentation $<X|R>$ is defined to be $|X|-|R|$. If $deg(G)>0$ then group $G$ is of order infinite.
answered Dec 7 '18 at 15:46
yavaryavar
843
answered Dec 7 '18 at 15:46
yavaryavar
843
answered Dec 7 '18 at 15:46
yavaryavar
843
answered Dec 7 '18 at 15:46
yavaryavar
843
843
add a comment |
add a comment |
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1
$begingroup$
If you add the relation $ab=ba$ then you quickly deduce $a=b$, so there is a surjective map to $mathbb Z$.
$endgroup$
– lulu
Dec 7 '18 at 13:13
$begingroup$
You can define your homomorphism to ${mathbb Z}$ by mapping $a$ and $b$ to $1$.
$endgroup$
– Derek Holt
Dec 7 '18 at 15:12