Showing $langle a,bmid abab^{-1}rangle$ and $ langle c,d mid c^2d^2rangle$ are isomorphic.
$begingroup$
I computed the fundamental group of the Klein bottle in two different ways and obtained two seemingly different answers:
$$
langle a,b mid abab^{-1}rangle
$$
and
$$
langle c,d mid c^2d^2rangle.
$$
I then tried to show that these two groups are in fact the same:
begin{align*}
langle a,b mid abab^{-1}rangle
=langle ab,b^{-1} mid abab^{-1}rangle
&=langle ab,b^{-1} mid (ab)(ab)b^{-1}b^{-1}rangle\
&=langle c,d mid c^2d^2rangle
end{align*}
Is my method for doing so valid?
group-theory proof-verification algebraic-topology fundamental-groups group-presentation
edited Dec 10 '18 at 6:04
Shaun
9,759113684
asked Dec 5 '18 at 21:53
ImNotThereRightNow_ImNotThereRightNow_
938
$endgroup$
add a comment |
$begingroup$
I computed the fundamental group of the Klein bottle in two different ways and obtained two seemingly different answers:
$$
langle a,b mid abab^{-1}rangle
$$
and
$$
langle c,d mid c^2d^2rangle.
$$
I then tried to show that these two groups are in fact the same:
begin{align*}
langle a,b mid abab^{-1}rangle
=langle ab,b^{-1} mid abab^{-1}rangle
&=langle ab,b^{-1} mid (ab)(ab)b^{-1}b^{-1}rangle\
&=langle c,d mid c^2d^2rangle
end{align*}
Is my method for doing so valid?
group-theory proof-verification algebraic-topology fundamental-groups group-presentation
edited Dec 10 '18 at 6:04
Shaun
9,759113684
asked Dec 5 '18 at 21:53
ImNotThereRightNow_ImNotThereRightNow_
938
$endgroup$
2
$begingroup$
Take a look at en.wikipedia.org/wiki/Tietze_transformations
$endgroup$
– Kyle Miller
Dec 6 '18 at 0:45
add a comment |
$begingroup$
I computed the fundamental group of the Klein bottle in two different ways and obtained two seemingly different answers:
$$
langle a,b mid abab^{-1}rangle
$$
and
$$
langle c,d mid c^2d^2rangle.
$$
I then tried to show that these two groups are in fact the same:
begin{align*}
langle a,b mid abab^{-1}rangle
=langle ab,b^{-1} mid abab^{-1}rangle
&=langle ab,b^{-1} mid (ab)(ab)b^{-1}b^{-1}rangle\
&=langle c,d mid c^2d^2rangle
end{align*}
Is my method for doing so valid?
group-theory proof-verification algebraic-topology fundamental-groups group-presentation
edited Dec 10 '18 at 6:04
Shaun
9,759113684
asked Dec 5 '18 at 21:53
ImNotThereRightNow_ImNotThereRightNow_
938
$endgroup$
I computed the fundamental group of the Klein bottle in two different ways and obtained two seemingly different answers:
$$
langle a,b mid abab^{-1}rangle
$$
and
$$
langle c,d mid c^2d^2rangle.
$$
I then tried to show that these two groups are in fact the same:
begin{align*}
langle a,b mid abab^{-1}rangle
=langle ab,b^{-1} mid abab^{-1}rangle
&=langle ab,b^{-1} mid (ab)(ab)b^{-1}b^{-1}rangle\
&=langle c,d mid c^2d^2rangle
end{align*}
Is my method for doing so valid?
group-theory proof-verification algebraic-topology fundamental-groups group-presentation
group-theory proof-verification algebraic-topology fundamental-groups group-presentation
edited Dec 10 '18 at 6:04
Shaun
9,759113684
asked Dec 5 '18 at 21:53
ImNotThereRightNow_ImNotThereRightNow_
938
edited Dec 10 '18 at 6:04
Shaun
9,759113684
asked Dec 5 '18 at 21:53
ImNotThereRightNow_ImNotThereRightNow_
938
edited Dec 10 '18 at 6:04
Shaun
9,759113684
edited Dec 10 '18 at 6:04
Shaun
9,759113684
edited Dec 10 '18 at 6:04
Shaun
9,759113684
9,759113684
asked Dec 5 '18 at 21:53
ImNotThereRightNow_ImNotThereRightNow_
938
asked Dec 5 '18 at 21:53
ImNotThereRightNow_ImNotThereRightNow_
938
asked Dec 5 '18 at 21:53
ImNotThereRightNow_ImNotThereRightNow_
938
938
2
$begingroup$
Take a look at en.wikipedia.org/wiki/Tietze_transformations
$endgroup$
– Kyle Miller
Dec 6 '18 at 0:45
add a comment |
2
$begingroup$
Take a look at en.wikipedia.org/wiki/Tietze_transformations
$endgroup$
– Kyle Miller
Dec 6 '18 at 0:45
2
2
$begingroup$
Take a look at en.wikipedia.org/wiki/Tietze_transformations
$endgroup$
– Kyle Miller
Dec 6 '18 at 0:45
$begingroup$
Take a look at en.wikipedia.org/wiki/Tietze_transformations
$endgroup$
– Kyle Miller
Dec 6 '18 at 0:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Sure. The second step doesn't really make sense as a presentation, but if you skip it then everything is fine. It might be clearest that these groups are isomorphic if you added the intermediate group $$langle a,b,c,d| abab^{-1},ab=c,d=b^{-1},c^2d^2rangle$$
which admits obvious maps from each of the two desired groups which one readily proves to be surjective (since $c$ and $d$ are in the subgroup generated by $a$ and $b$) and injective (since the first and last relations are each implied by the other three.)
answered Dec 5 '18 at 22:08
Kevin CarlsonKevin Carlson
33.6k23372
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Sure. The second step doesn't really make sense as a presentation, but if you skip it then everything is fine. It might be clearest that these groups are isomorphic if you added the intermediate group $$langle a,b,c,d| abab^{-1},ab=c,d=b^{-1},c^2d^2rangle$$
which admits obvious maps from each of the two desired groups which one readily proves to be surjective (since $c$ and $d$ are in the subgroup generated by $a$ and $b$) and injective (since the first and last relations are each implied by the other three.)
answered Dec 5 '18 at 22:08
Kevin CarlsonKevin Carlson
33.6k23372
$endgroup$
add a comment |
$begingroup$
Sure. The second step doesn't really make sense as a presentation, but if you skip it then everything is fine. It might be clearest that these groups are isomorphic if you added the intermediate group $$langle a,b,c,d| abab^{-1},ab=c,d=b^{-1},c^2d^2rangle$$
which admits obvious maps from each of the two desired groups which one readily proves to be surjective (since $c$ and $d$ are in the subgroup generated by $a$ and $b$) and injective (since the first and last relations are each implied by the other three.)
answered Dec 5 '18 at 22:08
Kevin CarlsonKevin Carlson
33.6k23372
$endgroup$
add a comment |
$begingroup$
Sure. The second step doesn't really make sense as a presentation, but if you skip it then everything is fine. It might be clearest that these groups are isomorphic if you added the intermediate group $$langle a,b,c,d| abab^{-1},ab=c,d=b^{-1},c^2d^2rangle$$
which admits obvious maps from each of the two desired groups which one readily proves to be surjective (since $c$ and $d$ are in the subgroup generated by $a$ and $b$) and injective (since the first and last relations are each implied by the other three.)
answered Dec 5 '18 at 22:08
Kevin CarlsonKevin Carlson
33.6k23372
$endgroup$
Sure. The second step doesn't really make sense as a presentation, but if you skip it then everything is fine. It might be clearest that these groups are isomorphic if you added the intermediate group $$langle a,b,c,d| abab^{-1},ab=c,d=b^{-1},c^2d^2rangle$$
which admits obvious maps from each of the two desired groups which one readily proves to be surjective (since $c$ and $d$ are in the subgroup generated by $a$ and $b$) and injective (since the first and last relations are each implied by the other three.)
answered Dec 5 '18 at 22:08
Kevin CarlsonKevin Carlson
33.6k23372
answered Dec 5 '18 at 22:08
Kevin CarlsonKevin Carlson
33.6k23372
answered Dec 5 '18 at 22:08
Kevin CarlsonKevin Carlson
33.6k23372
answered Dec 5 '18 at 22:08
Kevin CarlsonKevin Carlson
33.6k23372
33.6k23372
add a comment |
add a comment |
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$begingroup$
Take a look at en.wikipedia.org/wiki/Tietze_transformations
$endgroup$
– Kyle Miller
Dec 6 '18 at 0:45