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How many elements in the cyclic group $C_54$ have order $6$? How many have order $8$?
If a cyclic group has an element of infinite order, how many elements of finite order does it have?Abelian group with cyclic subgroup and cyclic quotient is generated by two elementsHow to find a generator of a cyclic group?Why is the set of integers with the operation of addition considered a cyclic group?A prime order group must be cyclicCan $S_n$ be a cyclic group?$(G:H) < infty$ where $G$ is an infinite cyclic groupHow is “integers mod n under addition” a cyclic group?Subgroup of a Dihedral group is Cyclic proofGroup theory problem - the order of elements
$begingroup$
I'm new to the algebra game and I'm learning about group theory at Uni. I understand a cyclic group is a group that can be generated by a single element, but I'm not sure how I would answer the above question.
abstract-algebra group-theory cyclic-groups
$endgroup$
|
show 2 more comments
$begingroup$
I'm new to the algebra game and I'm learning about group theory at Uni. I understand a cyclic group is a group that can be generated by a single element, but I'm not sure how I would answer the above question.
abstract-algebra group-theory cyclic-groups
$endgroup$
1
$begingroup$
Have you come across Lagrange's theorem? This will let you answer one of the two questions very quickly.
$endgroup$
– Arthur
Mar 21 at 12:18
$begingroup$
@Arthur yep the one which states ..every subgroup H of G divides the order of G, thank you for the help.
$endgroup$
– rishi547
Mar 21 at 12:27
$begingroup$
@DietrichBurde Ive looked mate, but most of it goes right over my head, thanks for the help though.
$endgroup$
– rishi547
Mar 21 at 12:28
$begingroup$
Hi Rishi, if an element has order $8$, then the group generated by that element must have order $8$ and be a subgroup of $C_54$, right? Now, by Lagrange's theorem which you correctly stated, $8$ should be a multiple of $54$. Is this true? What can you conclude?
$endgroup$
– астон вілла олоф мэллбэрг
Mar 21 at 12:55
$begingroup$
And for $n=6$, the subgroup generated by such an element is isomorphic to $C_6$, and this happens. Then there are $phi(6)=2$ generators.
$endgroup$
– Dietrich Burde
Mar 21 at 14:41
|
show 2 more comments
$begingroup$
I'm new to the algebra game and I'm learning about group theory at Uni. I understand a cyclic group is a group that can be generated by a single element, but I'm not sure how I would answer the above question.
abstract-algebra group-theory cyclic-groups
$endgroup$
I'm new to the algebra game and I'm learning about group theory at Uni. I understand a cyclic group is a group that can be generated by a single element, but I'm not sure how I would answer the above question.
abstract-algebra group-theory cyclic-groups
abstract-algebra group-theory cyclic-groups
edited Mar 21 at 12:37
Alan Muniz
2,61311030
2,61311030
asked Mar 21 at 12:16
rishi547rishi547
1
1
1
$begingroup$
Have you come across Lagrange's theorem? This will let you answer one of the two questions very quickly.
$endgroup$
– Arthur
Mar 21 at 12:18
$begingroup$
@Arthur yep the one which states ..every subgroup H of G divides the order of G, thank you for the help.
$endgroup$
– rishi547
Mar 21 at 12:27
$begingroup$
@DietrichBurde Ive looked mate, but most of it goes right over my head, thanks for the help though.
$endgroup$
– rishi547
Mar 21 at 12:28
$begingroup$
Hi Rishi, if an element has order $8$, then the group generated by that element must have order $8$ and be a subgroup of $C_54$, right? Now, by Lagrange's theorem which you correctly stated, $8$ should be a multiple of $54$. Is this true? What can you conclude?
$endgroup$
– астон вілла олоф мэллбэрг
Mar 21 at 12:55
$begingroup$
And for $n=6$, the subgroup generated by such an element is isomorphic to $C_6$, and this happens. Then there are $phi(6)=2$ generators.
$endgroup$
– Dietrich Burde
Mar 21 at 14:41
|
show 2 more comments
1
$begingroup$
Have you come across Lagrange's theorem? This will let you answer one of the two questions very quickly.
$endgroup$
– Arthur
Mar 21 at 12:18
$begingroup$
@Arthur yep the one which states ..every subgroup H of G divides the order of G, thank you for the help.
$endgroup$
– rishi547
Mar 21 at 12:27
$begingroup$
@DietrichBurde Ive looked mate, but most of it goes right over my head, thanks for the help though.
$endgroup$
– rishi547
Mar 21 at 12:28
$begingroup$
Hi Rishi, if an element has order $8$, then the group generated by that element must have order $8$ and be a subgroup of $C_54$, right? Now, by Lagrange's theorem which you correctly stated, $8$ should be a multiple of $54$. Is this true? What can you conclude?
$endgroup$
– астон вілла олоф мэллбэрг
Mar 21 at 12:55
$begingroup$
And for $n=6$, the subgroup generated by such an element is isomorphic to $C_6$, and this happens. Then there are $phi(6)=2$ generators.
$endgroup$
– Dietrich Burde
Mar 21 at 14:41
1
1
$begingroup$
Have you come across Lagrange's theorem? This will let you answer one of the two questions very quickly.
$endgroup$
– Arthur
Mar 21 at 12:18
$begingroup$
Have you come across Lagrange's theorem? This will let you answer one of the two questions very quickly.
$endgroup$
– Arthur
Mar 21 at 12:18
$begingroup$
@Arthur yep the one which states ..every subgroup H of G divides the order of G, thank you for the help.
$endgroup$
– rishi547
Mar 21 at 12:27
$begingroup$
@Arthur yep the one which states ..every subgroup H of G divides the order of G, thank you for the help.
$endgroup$
– rishi547
Mar 21 at 12:27
$begingroup$
@DietrichBurde Ive looked mate, but most of it goes right over my head, thanks for the help though.
$endgroup$
– rishi547
Mar 21 at 12:28
$begingroup$
@DietrichBurde Ive looked mate, but most of it goes right over my head, thanks for the help though.
$endgroup$
– rishi547
Mar 21 at 12:28
$begingroup$
Hi Rishi, if an element has order $8$, then the group generated by that element must have order $8$ and be a subgroup of $C_54$, right? Now, by Lagrange's theorem which you correctly stated, $8$ should be a multiple of $54$. Is this true? What can you conclude?
$endgroup$
– астон вілла олоф мэллбэрг
Mar 21 at 12:55
$begingroup$
Hi Rishi, if an element has order $8$, then the group generated by that element must have order $8$ and be a subgroup of $C_54$, right? Now, by Lagrange's theorem which you correctly stated, $8$ should be a multiple of $54$. Is this true? What can you conclude?
$endgroup$
– астон вілла олоф мэллбэрг
Mar 21 at 12:55
$begingroup$
And for $n=6$, the subgroup generated by such an element is isomorphic to $C_6$, and this happens. Then there are $phi(6)=2$ generators.
$endgroup$
– Dietrich Burde
Mar 21 at 14:41
$begingroup$
And for $n=6$, the subgroup generated by such an element is isomorphic to $C_6$, and this happens. Then there are $phi(6)=2$ generators.
$endgroup$
– Dietrich Burde
Mar 21 at 14:41
|
show 2 more comments
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1
$begingroup$
Have you come across Lagrange's theorem? This will let you answer one of the two questions very quickly.
$endgroup$
– Arthur
Mar 21 at 12:18
$begingroup$
@Arthur yep the one which states ..every subgroup H of G divides the order of G, thank you for the help.
$endgroup$
– rishi547
Mar 21 at 12:27
$begingroup$
@DietrichBurde Ive looked mate, but most of it goes right over my head, thanks for the help though.
$endgroup$
– rishi547
Mar 21 at 12:28
$begingroup$
Hi Rishi, if an element has order $8$, then the group generated by that element must have order $8$ and be a subgroup of $C_54$, right? Now, by Lagrange's theorem which you correctly stated, $8$ should be a multiple of $54$. Is this true? What can you conclude?
$endgroup$
– астон вілла олоф мэллбэрг
Mar 21 at 12:55
$begingroup$
And for $n=6$, the subgroup generated by such an element is isomorphic to $C_6$, and this happens. Then there are $phi(6)=2$ generators.
$endgroup$
– Dietrich Burde
Mar 21 at 14:41