Re: [math-fun] Simple finite group problem
Dan Asimov <asimov@msri.org> wrote:
Keith F. Lynch <kfl@KeithLynch.net> wrote:
List all multiplicative groups of integers mod ten. For instance {1,9} is one of them, and {1} is another.
If I understand the problem, it's to list all subgroups of the group of invertible elements of the ring Z/10.
The invertible elements can be chosen as {-3, -1, 1, 3}.
Since there is an element of order 4, this is isomorphic to the abelian group Z/4, so it has: ....
{-3, -1, 1, 3}, {-1, 1}, {1}.
Not quite what I meant. Yes, those are solutions, but those are not the only solutions. I meant any subset of {0,1,2,3,4,5,6,7,8,9} which forms a group under multiplication modulo 10. There are of course 1024 subsets, and I brute-force tested every one of them just for lulz. Similarly with all other moduli 1 through 36. How can I phrase that better? Thanks.
Modulo 10, only 1, 3, 7, and 9 have reciprocals. These four form a cyclic group under multiplication, generated by 3 (or 7). This group has only two subgroups, {1} and {1,9}. These are the three groups Dan enumerated. I don't think there are any others; so perhaps I still haven't understood what Keith is trying to enumerate. Keith, can you give another example of a subset of the integers modulo 10 that meets your criteria? On Wed, Apr 20, 2016 at 8:39 PM, Keith F. Lynch <kfl@keithlynch.net> wrote:
Dan Asimov <asimov@msri.org> wrote:
Keith F. Lynch <kfl@KeithLynch.net> wrote:
List all multiplicative groups of integers mod ten. For instance {1,9} is one of them, and {1} is another.
If I understand the problem, it's to list all subgroups of the group of invertible elements of the ring Z/10.
The invertible elements can be chosen as {-3, -1, 1, 3}.
Since there is an element of order 4, this is isomorphic to the abelian group Z/4, so it has: ....
{-3, -1, 1, 3}, {-1, 1}, {1}.
Not quite what I meant. Yes, those are solutions, but those are not the only solutions.
I meant any subset of {0,1,2,3,4,5,6,7,8,9} which forms a group under multiplication modulo 10. There are of course 1024 subsets, and I brute-force tested every one of them just for lulz. Similarly with all other moduli 1 through 36.
How can I phrase that better? Thanks.
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Wait, I think I just realized. Consider, for instance, {2,4,6,8}. Under multiplication, these do form a group with identity 6. I didn't think of it because it doesn't "cohere" with the ring of integers mod 10. {5} is another example. I'm not sure how to enumerate these. {6} and {4,6} are two more. On Wed, Apr 20, 2016 at 9:24 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Modulo 10, only 1, 3, 7, and 9 have reciprocals. These four form a cyclic group under multiplication, generated by 3 (or 7). This group has only two subgroups, {1} and {1,9}. These are the three groups Dan enumerated. I don't think there are any others; so perhaps I still haven't understood what Keith is trying to enumerate. Keith, can you give another example of a subset of the integers modulo 10 that meets your criteria?
On Wed, Apr 20, 2016 at 8:39 PM, Keith F. Lynch <kfl@keithlynch.net> wrote:
Dan Asimov <asimov@msri.org> wrote:
Keith F. Lynch <kfl@KeithLynch.net> wrote:
List all multiplicative groups of integers mod ten. For instance {1,9} is one of them, and {1} is another.
If I understand the problem, it's to list all subgroups of the group of invertible elements of the ring Z/10.
The invertible elements can be chosen as {-3, -1, 1, 3}.
Since there is an element of order 4, this is isomorphic to the abelian group Z/4, so it has: ....
{-3, -1, 1, 3}, {-1, 1}, {1}.
Not quite what I meant. Yes, those are solutions, but those are not the only solutions.
I meant any subset of {0,1,2,3,4,5,6,7,8,9} which forms a group under multiplication modulo 10. There are of course 1024 subsets, and I brute-force tested every one of them just for lulz. Similarly with all other moduli 1 through 36.
How can I phrase that better? Thanks.
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On Wed, Apr 20, 2016 at 6:30 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Wait, I think I just realized. Consider, for instance, {2,4,6,8}. Under multiplication, these do form a group with identity 6.
The Chinese remainder theorem lets you express integers mod 10 as pairs of integers mod 2 and integers mod 5; these are all the pairs (0, x) where x is nonzero.
I didn't think of it because it doesn't "cohere" with the ring of integers mod 10. {5} is another example.
These are all the pairs of the form (x, 0) where x is nonzero.
I'm not sure how to enumerate these. {6} and {4,6} are two more.
{6} is {(0,1)}. Its dual is {(1,0)} = {1} {4, 6} is {(0,-1), (0,1)}. Its dual is also {1}, since -1 = 1 mod 2. The answer is all products of subgroups of Z/2Z with those of Z/5Z. -- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
Fun! Let's see. We have: ---------------------------- {0}, {1}, {5}, {6} {-1, 1}, {4, 6}, {1, 5} {-1, 1, 5} {-3, -1, 1, 3}, {-4, -1, 1, 4}, {-4, -2, 2, 4} {-3, -1, 1, 3, 5} {-4, -3, -2, -1, 1, 2, 3, 4} ---------------------------- Plus all the rest. —Dan
On Apr 20, 2016, at 6:30 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Wait, I think I just realized. Consider, for instance, {2,4,6,8}. Under multiplication, these do form a group with identity 6. I didn't think of it because it doesn't "cohere" with the ring of integers mod 10. {5} is another example. I'm not sure how to enumerate these. {6} and {4,6} are two more.
Keith F. Lynch <kfl@KeithLynch.net <mailto:kfl@keithlynch.net>> wrote:
I meant any subset of {0,1,2,3,4,5,6,7,8,9} which forms a group under multiplication modulo 10.
Slow up! I don't think {1,5} is an example. Which is the identity? On Wed, Apr 20, 2016 at 10:09 PM, Dan Asimov <asimov@msri.org> wrote:
Fun!
Let's see. We have:
---------------------------- {0}, {1}, {5}, {6}
{-1, 1}, {4, 6}, {1, 5}
{-1, 1, 5}
{-3, -1, 1, 3}, {-4, -1, 1, 4}, {-4, -2, 2, 4}
{-3, -1, 1, 3, 5}
{-4, -3, -2, -1, 1, 2, 3, 4} ----------------------------
Plus all the rest.
—Dan
On Apr 20, 2016, at 6:30 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Wait, I think I just realized. Consider, for instance, {2,4,6,8}. Under multiplication, these do form a group with identity 6. I didn't think of it because it doesn't "cohere" with the ring of integers mod 10. {5} is another example. I'm not sure how to enumerate these. {6} and {4,6} are two more.
Keith F. Lynch <kfl@KeithLynch.net <mailto:kfl@keithlynch.net>> wrote:
I meant any subset of {0,1,2,3,4,5,6,7,8,9} which forms a group under multiplication modulo 10.
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1 is the identity. But you're right, 5 has no inverse. —Dan
On Apr 20, 2016, at 7:13 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Slow up! I don't think {1,5} is an example. Which is the identity?
On Wed, Apr 20, 2016 at 10:09 PM, Dan Asimov <asimov@msri.org> wrote:
Fun!
Let's see. We have:
---------------------------- {0}, {1}, {5}, {6}
{-1, 1}, {4, 6}, {1, 5}
{-1, 1, 5}
{-3, -1, 1, 3}, {-4, -1, 1, 4}, {-4, -2, 2, 4}
{-3, -1, 1, 3, 5}
{-4, -3, -2, -1, 1, 2, 3, 4} ----------------------------
Plus all the rest.
—Dan
On Apr 20, 2016, at 6:30 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Wait, I think I just realized. Consider, for instance, {2,4,6,8}. Under multiplication, these do form a group with identity 6. I didn't think of it because it doesn't "cohere" with the ring of integers mod 10. {5} is another example. I'm not sure how to enumerate these. {6} and {4,6} are two more.
Keith F. Lynch <kfl@KeithLynch.net <mailto:kfl@keithlynch.net>> wrote:
I meant any subset of {0,1,2,3,4,5,6,7,8,9} which forms a group under multiplication modulo 10.
math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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As Keith phrased it ----- any subset of {0,1,2,3,4,5,6,7,8,9} which forms a group under multiplication modulo 10 ---- , I think it would also fair to say {0} is a group under multiplication modulo 10. —Dan
On Apr 20, 2016, at 6:24 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Modulo 10, only 1, 3, 7, and 9 have reciprocals. These four form a cyclic group under multiplication, generated by 3 (or 7). This group has only two subgroups, {1} and {1,9}. These are the three groups Dan enumerated. I don't think there are any others; so perhaps I still haven't understood what Keith is trying to enumerate. Keith, can you give another example of a subset of the integers modulo 10 that meets your criteria?
On Wed, Apr 20, 2016 at 8:39 PM, Keith F. Lynch <kfl@keithlynch.net> wrote:
Dan Asimov <asimov@msri.org> wrote:
Keith F. Lynch <kfl@KeithLynch.net> wrote:
List all multiplicative groups of integers mod ten. For instance {1,9} is one of them, and {1} is another.
If I understand the problem, it's to list all subgroups of the group of invertible elements of the ring Z/10.
The invertible elements can be chosen as {-3, -1, 1, 3}.
Since there is an element of order 4, this is isomorphic to the abelian group Z/4, so it has: ....
{-3, -1, 1, 3}, {-1, 1}, {1}.
Not quite what I meant. Yes, those are solutions, but those are not the only solutions.
I meant any subset of {0,1,2,3,4,5,6,7,8,9} which forms a group under multiplication modulo 10. There are of course 1024 subsets, and I brute-force tested every one of them just for lulz. Similarly with all other moduli 1 through 36.
How can I phrase that better? Thanks.
Perhaps you want to emphasize that you are counting subsemigroups of the multiplicative semigroup Z_10 = {0,...,9} that are groups (under multiplication modulo 10) . If indeed that's what you are doing. If so, this will include for example all of the singleton sets containing only an idempotent, namely {0}, {1}, {5} and {6}. For every idempotent e (that is e^2 = e) in a semigroup S there is a maximal subsemigroup H(e) which is a group with identity e. Some papers that discuss the multiplicative group of integers mod n can be found in the following links: http://projecteuclid.org/download/pdf_1/euclid.pjm/1103038067 https://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n Something very close to your questions is considered in this paper Multiplicative Subgroups of Z/nZ <http://math.ucdenver.edu/~spayne/classnotes/subgroup.ps> Actually Z_n under multiplication is often called a commutative monoid since of course the multiplication is commutative and 1 acts as an identity. So this will give another search term. On Wed, Apr 20, 2016 at 9:33 PM, Dan Asimov <dasimov@earthlink.net> wrote:
As Keith phrased it
----- any subset of {0,1,2,3,4,5,6,7,8,9} which forms a group under multiplication modulo 10 ----
, I think it would also fair to say {0} is a group under multiplication modulo 10.
—Dan
On Apr 20, 2016, at 6:24 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Modulo 10, only 1, 3, 7, and 9 have reciprocals. These four form a cyclic group under multiplication, generated by 3 (or 7). This group has only two subgroups, {1} and {1,9}. These are the three groups Dan enumerated. I don't think there are any others; so perhaps I still haven't understood what Keith is trying to enumerate. Keith, can you give another example of a subset of the integers modulo 10 that meets your criteria?
On Wed, Apr 20, 2016 at 8:39 PM, Keith F. Lynch <kfl@keithlynch.net> wrote:
Dan Asimov <asimov@msri.org> wrote:
Keith F. Lynch <kfl@KeithLynch.net> wrote:
List all multiplicative groups of integers mod ten. For instance {1,9} is one of them, and {1} is another.
If I understand the problem, it's to list all subgroups of the group of invertible elements of the ring Z/10.
The invertible elements can be chosen as {-3, -1, 1, 3}.
Since there is an element of order 4, this is isomorphic to the abelian group Z/4, so it has: ....
{-3, -1, 1, 3}, {-1, 1}, {1}.
Not quite what I meant. Yes, those are solutions, but those are not the only solutions.
I meant any subset of {0,1,2,3,4,5,6,7,8,9} which forms a group under multiplication modulo 10. There are of course 1024 subsets, and I brute-force tested every one of them just for lulz. Similarly with all other moduli 1 through 36.
How can I phrase that better? Thanks.
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participants (6)
-
Allan Wechsler -
Dan Asimov -
Dan Asimov -
Keith F. Lynch -
Mike Stay -
W. Edwin Clark