Mixing
Any group of order four is either cyclic or isomorphic to $V$
$begingroup$
I will try and show that if our group $G$ is not cyclic then it is isomorphic to Klein's four-group $V$.
So suppose $G$ is not cyclic. Then its elements, excluding $e$, are of order $3$ or $2$. Supposing there is an element of order $3$ gives three distinct elements $g, g^2, g^3$. By hypothesis there exists $h in G$ such that $h$ is distinct from these three elements. So we can make $gh$ which must be equal to one of the elements previously mentioned. Setting this equal to the various elements:
begin{align}
&gh=g Rightarrow h=e && gh=g^2 Rightarrow h=g & \
&gh=h Rightarrow g=e, && gh=g^3 Rightarrow h=g^2 &
end{align}
Each of which is a contradiction.
Therefore all the elements are of order two and we can get an isomorphism between $G$ and $H times H$ where $ H$ is cyclic of order two. But this direct product is isomorphic to Klein's four-group so $G$ is isomorphic to $V$.
Is this correct?
Also it is just given in the book that $V$ is isomorphic to this product; how do I show this without checking that $f(gh)=f(g)f(h)$ for all combinations?
group-theory finite-groups
edited Jul 1 '12 at 19:59
Dylan Moreland
16.9k23564
asked Jul 1 '12 at 15:29
Ben DavidsonBen Davidson
3113
$endgroup$
add a comment |
$begingroup$
I will try and show that if our group $G$ is not cyclic then it is isomorphic to Klein's four-group $V$.
So suppose $G$ is not cyclic. Then its elements, excluding $e$, are of order $3$ or $2$. Supposing there is an element of order $3$ gives three distinct elements $g, g^2, g^3$. By hypothesis there exists $h in G$ such that $h$ is distinct from these three elements. So we can make $gh$ which must be equal to one of the elements previously mentioned. Setting this equal to the various elements:
begin{align}
&gh=g Rightarrow h=e && gh=g^2 Rightarrow h=g & \
&gh=h Rightarrow g=e, && gh=g^3 Rightarrow h=g^2 &
end{align}
Each of which is a contradiction.
Therefore all the elements are of order two and we can get an isomorphism between $G$ and $H times H$ where $ H$ is cyclic of order two. But this direct product is isomorphic to Klein's four-group so $G$ is isomorphic to $V$.
Is this correct?
Also it is just given in the book that $V$ is isomorphic to this product; how do I show this without checking that $f(gh)=f(g)f(h)$ for all combinations?
group-theory finite-groups
edited Jul 1 '12 at 19:59
Dylan Moreland
16.9k23564
asked Jul 1 '12 at 15:29
Ben DavidsonBen Davidson
3113
$endgroup$
2
$begingroup$
Given the detail you went into on showing there can be no element of order 3 (rather than just citing 'the order of an element must divide the order of the group'), simply saying 'we can get an isomorphism between G and $mathbb{Z}_2timesmathbb{Z}_2$' seems to be shortcutting a lot. I'd suggest simply building the multiplication table for the group following the basic logic 'consider the elements $f, g, h$ not the identity in $G$; then $f^2=g^2=h^2=e$ since they all must be of order 2. Now, $fg=ldots$' - once you've built it you can just compare it to the multiplication table for $V$.
$endgroup$
– Steven Stadnicki
Jul 1 '12 at 15:45
$begingroup$
There cannot be an element of order $3$ ($xneq e$).
$endgroup$
– mrs
Jul 1 '12 at 15:56
add a comment |
$begingroup$
I will try and show that if our group $G$ is not cyclic then it is isomorphic to Klein's four-group $V$.
So suppose $G$ is not cyclic. Then its elements, excluding $e$, are of order $3$ or $2$. Supposing there is an element of order $3$ gives three distinct elements $g, g^2, g^3$. By hypothesis there exists $h in G$ such that $h$ is distinct from these three elements. So we can make $gh$ which must be equal to one of the elements previously mentioned. Setting this equal to the various elements:
begin{align}
&gh=g Rightarrow h=e && gh=g^2 Rightarrow h=g & \
&gh=h Rightarrow g=e, && gh=g^3 Rightarrow h=g^2 &
end{align}
Each of which is a contradiction.
Therefore all the elements are of order two and we can get an isomorphism between $G$ and $H times H$ where $ H$ is cyclic of order two. But this direct product is isomorphic to Klein's four-group so $G$ is isomorphic to $V$.
Is this correct?
Also it is just given in the book that $V$ is isomorphic to this product; how do I show this without checking that $f(gh)=f(g)f(h)$ for all combinations?
group-theory finite-groups
edited Jul 1 '12 at 19:59
Dylan Moreland
16.9k23564
asked Jul 1 '12 at 15:29
Ben DavidsonBen Davidson
3113
$endgroup$
I will try and show that if our group $G$ is not cyclic then it is isomorphic to Klein's four-group $V$.
So suppose $G$ is not cyclic. Then its elements, excluding $e$, are of order $3$ or $2$. Supposing there is an element of order $3$ gives three distinct elements $g, g^2, g^3$. By hypothesis there exists $h in G$ such that $h$ is distinct from these three elements. So we can make $gh$ which must be equal to one of the elements previously mentioned. Setting this equal to the various elements:
begin{align}
&gh=g Rightarrow h=e && gh=g^2 Rightarrow h=g & \
&gh=h Rightarrow g=e, && gh=g^3 Rightarrow h=g^2 &
end{align}
Each of which is a contradiction.
Therefore all the elements are of order two and we can get an isomorphism between $G$ and $H times H$ where $ H$ is cyclic of order two. But this direct product is isomorphic to Klein's four-group so $G$ is isomorphic to $V$.
Is this correct?
Also it is just given in the book that $V$ is isomorphic to this product; how do I show this without checking that $f(gh)=f(g)f(h)$ for all combinations?
group-theory finite-groups
group-theory finite-groups
edited Jul 1 '12 at 19:59
Dylan Moreland
16.9k23564
asked Jul 1 '12 at 15:29
Ben DavidsonBen Davidson
3113
edited Jul 1 '12 at 19:59
Dylan Moreland
16.9k23564
asked Jul 1 '12 at 15:29
Ben DavidsonBen Davidson
3113
edited Jul 1 '12 at 19:59
Dylan Moreland
16.9k23564
edited Jul 1 '12 at 19:59
Dylan Moreland
16.9k23564
edited Jul 1 '12 at 19:59
Dylan Moreland
16.9k23564
16.9k23564
asked Jul 1 '12 at 15:29
Ben DavidsonBen Davidson
3113
asked Jul 1 '12 at 15:29
Ben DavidsonBen Davidson
3113
asked Jul 1 '12 at 15:29
Ben DavidsonBen Davidson
3113
3113
2
$begingroup$
Given the detail you went into on showing there can be no element of order 3 (rather than just citing 'the order of an element must divide the order of the group'), simply saying 'we can get an isomorphism between G and $mathbb{Z}_2timesmathbb{Z}_2$' seems to be shortcutting a lot. I'd suggest simply building the multiplication table for the group following the basic logic 'consider the elements $f, g, h$ not the identity in $G$; then $f^2=g^2=h^2=e$ since they all must be of order 2. Now, $fg=ldots$' - once you've built it you can just compare it to the multiplication table for $V$.
$endgroup$
– Steven Stadnicki
Jul 1 '12 at 15:45
$begingroup$
There cannot be an element of order $3$ ($xneq e$).
$endgroup$
– mrs
Jul 1 '12 at 15:56
add a comment |
2
$begingroup$
Given the detail you went into on showing there can be no element of order 3 (rather than just citing 'the order of an element must divide the order of the group'), simply saying 'we can get an isomorphism between G and $mathbb{Z}_2timesmathbb{Z}_2$' seems to be shortcutting a lot. I'd suggest simply building the multiplication table for the group following the basic logic 'consider the elements $f, g, h$ not the identity in $G$; then $f^2=g^2=h^2=e$ since they all must be of order 2. Now, $fg=ldots$' - once you've built it you can just compare it to the multiplication table for $V$.
$endgroup$
– Steven Stadnicki
Jul 1 '12 at 15:45
$begingroup$
There cannot be an element of order $3$ ($xneq e$).
$endgroup$
– mrs
Jul 1 '12 at 15:56
2
2
$begingroup$
Given the detail you went into on showing there can be no element of order 3 (rather than just citing 'the order of an element must divide the order of the group'), simply saying 'we can get an isomorphism between G and $mathbb{Z}_2timesmathbb{Z}_2$' seems to be shortcutting a lot. I'd suggest simply building the multiplication table for the group following the basic logic 'consider the elements $f, g, h$ not the identity in $G$; then $f^2=g^2=h^2=e$ since they all must be of order 2. Now, $fg=ldots$' - once you've built it you can just compare it to the multiplication table for $V$.
$endgroup$
– Steven Stadnicki
Jul 1 '12 at 15:45
$begingroup$
Given the detail you went into on showing there can be no element of order 3 (rather than just citing 'the order of an element must divide the order of the group'), simply saying 'we can get an isomorphism between G and $mathbb{Z}_2timesmathbb{Z}_2$' seems to be shortcutting a lot. I'd suggest simply building the multiplication table for the group following the basic logic 'consider the elements $f, g, h$ not the identity in $G$; then $f^2=g^2=h^2=e$ since they all must be of order 2. Now, $fg=ldots$' - once you've built it you can just compare it to the multiplication table for $V$.
$endgroup$
– Steven Stadnicki
Jul 1 '12 at 15:45
$begingroup$
There cannot be an element of order $3$ ($xneq e$).
$endgroup$
– mrs
Jul 1 '12 at 15:56
$begingroup$
There cannot be an element of order $3$ ($xneq e$).
$endgroup$
– mrs
Jul 1 '12 at 15:56
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
An idea: let $,G={1,a,b,c},$ be a non cyclic group of order $,4,$, so we can assume $,a,b,$ are not powers of each other.
The question is: what is $,ab,$?? You should find pretty easy to show this must be $,c,$ and, with a few lines more, you prove both that $,G,$ is abelian and in fact the same (i.e., isomorphic to) the Viergruppe $,V,$.
Another idea: you seem to feel pretty comfortable using isomorphisms, Klein group and stuff: why won't you use directly Lagrange's Theorem and avoid all that long proof about the impossibility of an element of order 3?
edited Jan 13 at 17:33
Parcly Taxel
41.8k1372101
answered Jul 1 '12 at 15:56
DonAntonioDonAntonio
178k1494230
$endgroup$
4
$begingroup$
Your first line... should that be "$a,b$ are not powers of each other"?
$endgroup$
– Arturo Magidin
Jul 1 '12 at 18:04
$begingroup$
Of course....yikes! Thank you @ArturoMagidin . Corrected
$endgroup$
– DonAntonio
Jul 1 '12 at 19:43
add a comment |
$begingroup$
I was going to ask if my proof of this is correct, so here it is, please tell me if it is fine.
Proof.
Let G be a group of order 4 by the Lagrange theorem any element of G has order 1,2 or 4. If G has an element of order 4 then it generates so G is cyclic. Assume G has not elements of order 4, then G has 3 elements of order 2. Since the order of G is $2^2$ then G is Abelian (by the class equation). So G is Abelian and has 3 elements of order 2 and 1 of order 1 (the identity), it follows that G is isomorphic to the klein four group.
answered Dec 14 '12 at 6:11
CybusterCybuster
13718
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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active
oldest
votes
$begingroup$
An idea: let $,G={1,a,b,c},$ be a non cyclic group of order $,4,$, so we can assume $,a,b,$ are not powers of each other.
The question is: what is $,ab,$?? You should find pretty easy to show this must be $,c,$ and, with a few lines more, you prove both that $,G,$ is abelian and in fact the same (i.e., isomorphic to) the Viergruppe $,V,$.
Another idea: you seem to feel pretty comfortable using isomorphisms, Klein group and stuff: why won't you use directly Lagrange's Theorem and avoid all that long proof about the impossibility of an element of order 3?
edited Jan 13 at 17:33
Parcly Taxel
41.8k1372101
answered Jul 1 '12 at 15:56
DonAntonioDonAntonio
178k1494230
$endgroup$
4
$begingroup$
Your first line... should that be "$a,b$ are not powers of each other"?
$endgroup$
– Arturo Magidin
Jul 1 '12 at 18:04
$begingroup$
Of course....yikes! Thank you @ArturoMagidin . Corrected
$endgroup$
– DonAntonio
Jul 1 '12 at 19:43
add a comment |
$begingroup$
An idea: let $,G={1,a,b,c},$ be a non cyclic group of order $,4,$, so we can assume $,a,b,$ are not powers of each other.
The question is: what is $,ab,$?? You should find pretty easy to show this must be $,c,$ and, with a few lines more, you prove both that $,G,$ is abelian and in fact the same (i.e., isomorphic to) the Viergruppe $,V,$.
Another idea: you seem to feel pretty comfortable using isomorphisms, Klein group and stuff: why won't you use directly Lagrange's Theorem and avoid all that long proof about the impossibility of an element of order 3?
edited Jan 13 at 17:33
Parcly Taxel
41.8k1372101
answered Jul 1 '12 at 15:56
DonAntonioDonAntonio
178k1494230
$endgroup$
4
$begingroup$
Your first line... should that be "$a,b$ are not powers of each other"?
$endgroup$
– Arturo Magidin
Jul 1 '12 at 18:04
$begingroup$
Of course....yikes! Thank you @ArturoMagidin . Corrected
$endgroup$
– DonAntonio
Jul 1 '12 at 19:43
add a comment |
$begingroup$
An idea: let $,G={1,a,b,c},$ be a non cyclic group of order $,4,$, so we can assume $,a,b,$ are not powers of each other.
The question is: what is $,ab,$?? You should find pretty easy to show this must be $,c,$ and, with a few lines more, you prove both that $,G,$ is abelian and in fact the same (i.e., isomorphic to) the Viergruppe $,V,$.
Another idea: you seem to feel pretty comfortable using isomorphisms, Klein group and stuff: why won't you use directly Lagrange's Theorem and avoid all that long proof about the impossibility of an element of order 3?
edited Jan 13 at 17:33
Parcly Taxel
41.8k1372101
answered Jul 1 '12 at 15:56
DonAntonioDonAntonio
178k1494230
$endgroup$
An idea: let $,G={1,a,b,c},$ be a non cyclic group of order $,4,$, so we can assume $,a,b,$ are not powers of each other.
The question is: what is $,ab,$?? You should find pretty easy to show this must be $,c,$ and, with a few lines more, you prove both that $,G,$ is abelian and in fact the same (i.e., isomorphic to) the Viergruppe $,V,$.
Another idea: you seem to feel pretty comfortable using isomorphisms, Klein group and stuff: why won't you use directly Lagrange's Theorem and avoid all that long proof about the impossibility of an element of order 3?
edited Jan 13 at 17:33
Parcly Taxel
41.8k1372101
answered Jul 1 '12 at 15:56
DonAntonioDonAntonio
178k1494230
edited Jan 13 at 17:33
Parcly Taxel
41.8k1372101
edited Jan 13 at 17:33
Parcly Taxel
41.8k1372101
edited Jan 13 at 17:33
Parcly Taxel
41.8k1372101
41.8k1372101
answered Jul 1 '12 at 15:56
DonAntonioDonAntonio
178k1494230
answered Jul 1 '12 at 15:56
DonAntonioDonAntonio
178k1494230
answered Jul 1 '12 at 15:56
DonAntonioDonAntonio
178k1494230
178k1494230
4
$begingroup$
Your first line... should that be "$a,b$ are not powers of each other"?
$endgroup$
– Arturo Magidin
Jul 1 '12 at 18:04
$begingroup$
Of course....yikes! Thank you @ArturoMagidin . Corrected
$endgroup$
– DonAntonio
Jul 1 '12 at 19:43
add a comment |
4
$begingroup$
Your first line... should that be "$a,b$ are not powers of each other"?
$endgroup$
– Arturo Magidin
Jul 1 '12 at 18:04
$begingroup$
Of course....yikes! Thank you @ArturoMagidin . Corrected
$endgroup$
– DonAntonio
Jul 1 '12 at 19:43
4
4
$begingroup$
Your first line... should that be "$a,b$ are not powers of each other"?
$endgroup$
– Arturo Magidin
Jul 1 '12 at 18:04
$begingroup$
Your first line... should that be "$a,b$ are not powers of each other"?
$endgroup$
– Arturo Magidin
Jul 1 '12 at 18:04
$begingroup$
Of course....yikes! Thank you @ArturoMagidin . Corrected
$endgroup$
– DonAntonio
Jul 1 '12 at 19:43
$begingroup$
Of course....yikes! Thank you @ArturoMagidin . Corrected
$endgroup$
– DonAntonio
Jul 1 '12 at 19:43
add a comment |
$begingroup$
I was going to ask if my proof of this is correct, so here it is, please tell me if it is fine.
Proof.
Let G be a group of order 4 by the Lagrange theorem any element of G has order 1,2 or 4. If G has an element of order 4 then it generates so G is cyclic. Assume G has not elements of order 4, then G has 3 elements of order 2. Since the order of G is $2^2$ then G is Abelian (by the class equation). So G is Abelian and has 3 elements of order 2 and 1 of order 1 (the identity), it follows that G is isomorphic to the klein four group.
answered Dec 14 '12 at 6:11
CybusterCybuster
13718
$endgroup$
add a comment |
$begingroup$
I was going to ask if my proof of this is correct, so here it is, please tell me if it is fine.
Proof.
Let G be a group of order 4 by the Lagrange theorem any element of G has order 1,2 or 4. If G has an element of order 4 then it generates so G is cyclic. Assume G has not elements of order 4, then G has 3 elements of order 2. Since the order of G is $2^2$ then G is Abelian (by the class equation). So G is Abelian and has 3 elements of order 2 and 1 of order 1 (the identity), it follows that G is isomorphic to the klein four group.
answered Dec 14 '12 at 6:11
CybusterCybuster
13718
$endgroup$
add a comment |
$begingroup$
I was going to ask if my proof of this is correct, so here it is, please tell me if it is fine.
Proof.
Let G be a group of order 4 by the Lagrange theorem any element of G has order 1,2 or 4. If G has an element of order 4 then it generates so G is cyclic. Assume G has not elements of order 4, then G has 3 elements of order 2. Since the order of G is $2^2$ then G is Abelian (by the class equation). So G is Abelian and has 3 elements of order 2 and 1 of order 1 (the identity), it follows that G is isomorphic to the klein four group.
answered Dec 14 '12 at 6:11
CybusterCybuster
13718
$endgroup$
I was going to ask if my proof of this is correct, so here it is, please tell me if it is fine.
Proof.
Let G be a group of order 4 by the Lagrange theorem any element of G has order 1,2 or 4. If G has an element of order 4 then it generates so G is cyclic. Assume G has not elements of order 4, then G has 3 elements of order 2. Since the order of G is $2^2$ then G is Abelian (by the class equation). So G is Abelian and has 3 elements of order 2 and 1 of order 1 (the identity), it follows that G is isomorphic to the klein four group.
answered Dec 14 '12 at 6:11
CybusterCybuster
13718
edited Dec 14 '12 at 6:19
answered Dec 14 '12 at 6:11
CybusterCybuster
13718
answered Dec 14 '12 at 6:11
CybusterCybuster
13718
answered Dec 14 '12 at 6:11
CybusterCybuster
13718
13718
add a comment |
add a comment |
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$begingroup$
Given the detail you went into on showing there can be no element of order 3 (rather than just citing 'the order of an element must divide the order of the group'), simply saying 'we can get an isomorphism between G and $mathbb{Z}_2timesmathbb{Z}_2$' seems to be shortcutting a lot. I'd suggest simply building the multiplication table for the group following the basic logic 'consider the elements $f, g, h$ not the identity in $G$; then $f^2=g^2=h^2=e$ since they all must be of order 2. Now, $fg=ldots$' - once you've built it you can just compare it to the multiplication table for $V$.
$endgroup$
– Steven Stadnicki
Jul 1 '12 at 15:45
$begingroup$
There cannot be an element of order $3$ ($xneq e$).
$endgroup$
– mrs
Jul 1 '12 at 15:56