Mixing
Smallest subgroup generated by a subset of a group.
$begingroup$
If$ G$ is a group, and the set $S={a,b}$ is a subset of $ G$, can we say that the smallest subgroup of $ G$ generated by $langle a,brangle$ will always be either $langle a rangle$, $langle brangle$, or in the case that $langle arangle$ does not generate $b$ and $langle brangle$ does not generate $a$, then $langle a,brangle$ = $G$?
I'm having difficulty trying to think of a counterexample to this, particularly a finite group whose elements can be easily enumerated (e.g. $ U(n)$ the multiplicative group of integers modulo $n$).
group-theory finite-groups examples-counterexamples cyclic-groups
edited Jan 30 at 17:46
Shaun
10.1k113685
asked Jan 30 at 2:21
Nicholas CousarNicholas Cousar
374212
$endgroup$
add a comment |
$begingroup$
If$ G$ is a group, and the set $S={a,b}$ is a subset of $ G$, can we say that the smallest subgroup of $ G$ generated by $langle a,brangle$ will always be either $langle a rangle$, $langle brangle$, or in the case that $langle arangle$ does not generate $b$ and $langle brangle$ does not generate $a$, then $langle a,brangle$ = $G$?
I'm having difficulty trying to think of a counterexample to this, particularly a finite group whose elements can be easily enumerated (e.g. $ U(n)$ the multiplicative group of integers modulo $n$).
group-theory finite-groups examples-counterexamples cyclic-groups
edited Jan 30 at 17:46
Shaun
10.1k113685
asked Jan 30 at 2:21
Nicholas CousarNicholas Cousar
374212
$endgroup$
add a comment |
$begingroup$
If$ G$ is a group, and the set $S={a,b}$ is a subset of $ G$, can we say that the smallest subgroup of $ G$ generated by $langle a,brangle$ will always be either $langle a rangle$, $langle brangle$, or in the case that $langle arangle$ does not generate $b$ and $langle brangle$ does not generate $a$, then $langle a,brangle$ = $G$?
I'm having difficulty trying to think of a counterexample to this, particularly a finite group whose elements can be easily enumerated (e.g. $ U(n)$ the multiplicative group of integers modulo $n$).
group-theory finite-groups examples-counterexamples cyclic-groups
edited Jan 30 at 17:46
Shaun
10.1k113685
asked Jan 30 at 2:21
Nicholas CousarNicholas Cousar
374212
$endgroup$
If$ G$ is a group, and the set $S={a,b}$ is a subset of $ G$, can we say that the smallest subgroup of $ G$ generated by $langle a,brangle$ will always be either $langle a rangle$, $langle brangle$, or in the case that $langle arangle$ does not generate $b$ and $langle brangle$ does not generate $a$, then $langle a,brangle$ = $G$?
I'm having difficulty trying to think of a counterexample to this, particularly a finite group whose elements can be easily enumerated (e.g. $ U(n)$ the multiplicative group of integers modulo $n$).
group-theory finite-groups examples-counterexamples cyclic-groups
group-theory finite-groups examples-counterexamples cyclic-groups
edited Jan 30 at 17:46
Shaun
10.1k113685
asked Jan 30 at 2:21
Nicholas CousarNicholas Cousar
374212
edited Jan 30 at 17:46
Shaun
10.1k113685
asked Jan 30 at 2:21
Nicholas CousarNicholas Cousar
374212
edited Jan 30 at 17:46
Shaun
10.1k113685
edited Jan 30 at 17:46
Shaun
10.1k113685
edited Jan 30 at 17:46
Shaun
10.1k113685
10.1k113685
asked Jan 30 at 2:21
Nicholas CousarNicholas Cousar
374212
asked Jan 30 at 2:21
Nicholas CousarNicholas Cousar
374212
asked Jan 30 at 2:21
Nicholas CousarNicholas Cousar
374212
374212
add a comment |
add a comment |
1 Answer
1
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$begingroup$
No. Just take the group $C_2 times C_2 times C_2$ to see this. To motivate a little the construction of an example of a group which fails to satisfy your property, think of the situation like this:
If we can find a proper subgroup $H$ of $G$ such that $H$ is not cyclic, $G$ will definitely fail to satisfy your property. Do you see why?
Then it is easy to see that the smallest non-cyclic group is $V = C_2 times C_2$. How can we have $V$ as a proper subgroup of some other group? The most obvious way to do that is to take $G = A times V$, where $A>1$ is any group. The example I gave is the choice $A = C_2$ (and is a group of smallest possible order which fails to satisfy your property).
answered Jan 30 at 3:16
the_foxthe_fox
2,90231538
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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votes
$begingroup$
No. Just take the group $C_2 times C_2 times C_2$ to see this. To motivate a little the construction of an example of a group which fails to satisfy your property, think of the situation like this:
If we can find a proper subgroup $H$ of $G$ such that $H$ is not cyclic, $G$ will definitely fail to satisfy your property. Do you see why?
Then it is easy to see that the smallest non-cyclic group is $V = C_2 times C_2$. How can we have $V$ as a proper subgroup of some other group? The most obvious way to do that is to take $G = A times V$, where $A>1$ is any group. The example I gave is the choice $A = C_2$ (and is a group of smallest possible order which fails to satisfy your property).
answered Jan 30 at 3:16
the_foxthe_fox
2,90231538
$endgroup$
add a comment |
$begingroup$
No. Just take the group $C_2 times C_2 times C_2$ to see this. To motivate a little the construction of an example of a group which fails to satisfy your property, think of the situation like this:
If we can find a proper subgroup $H$ of $G$ such that $H$ is not cyclic, $G$ will definitely fail to satisfy your property. Do you see why?
Then it is easy to see that the smallest non-cyclic group is $V = C_2 times C_2$. How can we have $V$ as a proper subgroup of some other group? The most obvious way to do that is to take $G = A times V$, where $A>1$ is any group. The example I gave is the choice $A = C_2$ (and is a group of smallest possible order which fails to satisfy your property).
answered Jan 30 at 3:16
the_foxthe_fox
2,90231538
$endgroup$
add a comment |
$begingroup$
No. Just take the group $C_2 times C_2 times C_2$ to see this. To motivate a little the construction of an example of a group which fails to satisfy your property, think of the situation like this:
If we can find a proper subgroup $H$ of $G$ such that $H$ is not cyclic, $G$ will definitely fail to satisfy your property. Do you see why?
Then it is easy to see that the smallest non-cyclic group is $V = C_2 times C_2$. How can we have $V$ as a proper subgroup of some other group? The most obvious way to do that is to take $G = A times V$, where $A>1$ is any group. The example I gave is the choice $A = C_2$ (and is a group of smallest possible order which fails to satisfy your property).
answered Jan 30 at 3:16
the_foxthe_fox
2,90231538
$endgroup$
No. Just take the group $C_2 times C_2 times C_2$ to see this. To motivate a little the construction of an example of a group which fails to satisfy your property, think of the situation like this:
If we can find a proper subgroup $H$ of $G$ such that $H$ is not cyclic, $G$ will definitely fail to satisfy your property. Do you see why?
Then it is easy to see that the smallest non-cyclic group is $V = C_2 times C_2$. How can we have $V$ as a proper subgroup of some other group? The most obvious way to do that is to take $G = A times V$, where $A>1$ is any group. The example I gave is the choice $A = C_2$ (and is a group of smallest possible order which fails to satisfy your property).
answered Jan 30 at 3:16
the_foxthe_fox
2,90231538
edited Jan 30 at 16:57
answered Jan 30 at 3:16
the_foxthe_fox
2,90231538
answered Jan 30 at 3:16
the_foxthe_fox
2,90231538
answered Jan 30 at 3:16
the_foxthe_fox
2,90231538
2,90231538
add a comment |
add a comment |
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