Mixing
Sufficient conditions on the isomorphism of two groups
$begingroup$
Let $G_1, G_2$ be two groups with at least one nontrivial proper subgroup each.
Let $S_1, S_2$ be the sets of proper subgroups of, respectively $G_1, G_2$.
Suppose there exists a bijective function $f: S_1 rightarrow S_2$ such that $forall Ain S_1, f(A)$ is isomorphic to $A$.
When can I conclude that $G_1, G_2$ are isomorphic?
I think that, if $G_1$ and $G_2$ are finite and abelian we can conclude that they are isomorphic, but I can't prove It.
Moreover, I haven't found any counterexample for nonabelian finite groups.
group-theory
asked Jan 28 at 17:35
Lucio TanziniLucio Tanzini
351114
$endgroup$
add a comment |
$begingroup$
Let $G_1, G_2$ be two groups with at least one nontrivial proper subgroup each.
Let $S_1, S_2$ be the sets of proper subgroups of, respectively $G_1, G_2$.
Suppose there exists a bijective function $f: S_1 rightarrow S_2$ such that $forall Ain S_1, f(A)$ is isomorphic to $A$.
When can I conclude that $G_1, G_2$ are isomorphic?
I think that, if $G_1$ and $G_2$ are finite and abelian we can conclude that they are isomorphic, but I can't prove It.
Moreover, I haven't found any counterexample for nonabelian finite groups.
group-theory
asked Jan 28 at 17:35
Lucio TanziniLucio Tanzini
351114
$endgroup$
$begingroup$
$f(A)$ is a subgroup
$endgroup$
– the_fox
Jan 28 at 17:53
1
$begingroup$
If the two groups are not finite we surely can't conclude anything. A counterexample is $mathbb{Z}_2 times mathbb{Z}_4 times mathbb{Z}_4...$ and $mathbb{Z}_4 times mathbb{Z}_4...$
$endgroup$
– Lucio Tanzini
Jan 28 at 17:54
1
$begingroup$
@the_fox Ah, you are right, sets of proper subgroups, this was missing. I am sorry. Still, the question is a bit vague "when can I conclude that $G_1cong G_2$." Certainly not always, but sometimes.
$endgroup$
– Dietrich Burde
Jan 28 at 17:56
add a comment |
$begingroup$
Let $G_1, G_2$ be two groups with at least one nontrivial proper subgroup each.
Let $S_1, S_2$ be the sets of proper subgroups of, respectively $G_1, G_2$.
Suppose there exists a bijective function $f: S_1 rightarrow S_2$ such that $forall Ain S_1, f(A)$ is isomorphic to $A$.
When can I conclude that $G_1, G_2$ are isomorphic?
I think that, if $G_1$ and $G_2$ are finite and abelian we can conclude that they are isomorphic, but I can't prove It.
Moreover, I haven't found any counterexample for nonabelian finite groups.
group-theory
asked Jan 28 at 17:35
Lucio TanziniLucio Tanzini
351114
$endgroup$
Let $G_1, G_2$ be two groups with at least one nontrivial proper subgroup each.
Let $S_1, S_2$ be the sets of proper subgroups of, respectively $G_1, G_2$.
Suppose there exists a bijective function $f: S_1 rightarrow S_2$ such that $forall Ain S_1, f(A)$ is isomorphic to $A$.
When can I conclude that $G_1, G_2$ are isomorphic?
I think that, if $G_1$ and $G_2$ are finite and abelian we can conclude that they are isomorphic, but I can't prove It.
Moreover, I haven't found any counterexample for nonabelian finite groups.
group-theory
group-theory
asked Jan 28 at 17:35
Lucio TanziniLucio Tanzini
351114
asked Jan 28 at 17:35
Lucio TanziniLucio Tanzini
351114
edited Jan 28 at 21:46
Lucio Tanzini
asked Jan 28 at 17:35
Lucio TanziniLucio Tanzini
351114
asked Jan 28 at 17:35
Lucio TanziniLucio Tanzini
351114
asked Jan 28 at 17:35
Lucio TanziniLucio Tanzini
351114
351114
$begingroup$
$f(A)$ is a subgroup
$endgroup$
– the_fox
Jan 28 at 17:53
1
$begingroup$
If the two groups are not finite we surely can't conclude anything. A counterexample is $mathbb{Z}_2 times mathbb{Z}_4 times mathbb{Z}_4...$ and $mathbb{Z}_4 times mathbb{Z}_4...$
$endgroup$
– Lucio Tanzini
Jan 28 at 17:54
1
$begingroup$
@the_fox Ah, you are right, sets of proper subgroups, this was missing. I am sorry. Still, the question is a bit vague "when can I conclude that $G_1cong G_2$." Certainly not always, but sometimes.
$endgroup$
– Dietrich Burde
Jan 28 at 17:56
add a comment |
$begingroup$
$f(A)$ is a subgroup
$endgroup$
– the_fox
Jan 28 at 17:53
1
$begingroup$
If the two groups are not finite we surely can't conclude anything. A counterexample is $mathbb{Z}_2 times mathbb{Z}_4 times mathbb{Z}_4...$ and $mathbb{Z}_4 times mathbb{Z}_4...$
$endgroup$
– Lucio Tanzini
Jan 28 at 17:54
1
$begingroup$
@the_fox Ah, you are right, sets of proper subgroups, this was missing. I am sorry. Still, the question is a bit vague "when can I conclude that $G_1cong G_2$." Certainly not always, but sometimes.
$endgroup$
– Dietrich Burde
Jan 28 at 17:56
$begingroup$
$f(A)$ is a subgroup
$endgroup$
– the_fox
Jan 28 at 17:53
$begingroup$
$f(A)$ is a subgroup
$endgroup$
– the_fox
Jan 28 at 17:53
1
1
$begingroup$
If the two groups are not finite we surely can't conclude anything. A counterexample is $mathbb{Z}_2 times mathbb{Z}_4 times mathbb{Z}_4...$ and $mathbb{Z}_4 times mathbb{Z}_4...$
$endgroup$
– Lucio Tanzini
Jan 28 at 17:54
$begingroup$
If the two groups are not finite we surely can't conclude anything. A counterexample is $mathbb{Z}_2 times mathbb{Z}_4 times mathbb{Z}_4...$ and $mathbb{Z}_4 times mathbb{Z}_4...$
$endgroup$
– Lucio Tanzini
Jan 28 at 17:54
1
1
$begingroup$
@the_fox Ah, you are right, sets of proper subgroups, this was missing. I am sorry. Still, the question is a bit vague "when can I conclude that $G_1cong G_2$." Certainly not always, but sometimes.
$endgroup$
– Dietrich Burde
Jan 28 at 17:56
$begingroup$
@the_fox Ah, you are right, sets of proper subgroups, this was missing. I am sorry. Still, the question is a bit vague "when can I conclude that $G_1cong G_2$." Certainly not always, but sometimes.
$endgroup$
– Dietrich Burde
Jan 28 at 17:56
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
There are two pairs of examples of order $16$. These are the smallest examples. One of these two pairs is $C_4times C_4$ and $C_4rtimes C_4$. For both of these, the complete list of proper subgroups is:
1 trivial subgroup
3 subgroups isomorphic to $C_2$
6 subgroups isomorphic to $C_4$
1 subgroup isomorphic to $C_2times C_2$
3 subgroups isomorphic to $C_4times C_2$
(See https://groupprops.subwiki.org/wiki/Nontrivial_semidirect_product_of_Z4_and_Z4#Subgroups for the subgroups of $C_4rtimes C_4$.)
Another easy pair of examples is $C_9times C_3$ and $C_9rtimes C_3$.
(It is definitely true for finite abelian groups though, this is an easy consequence of their classification.)
answered Jan 28 at 22:07
verretverret
3,3171923
$endgroup$
add a comment |
$begingroup$
Certainly not always. I'd be surprised if there is a concrete set of conditions which is both necessary and sufficient to conclude isomorphism between the two groups. (My answer refers to finite groups only.)
There are groups which are called $P$-groups in Schmidt's book "Subgroup Lattices of Groups" (not be confused with $p$-groups) and which are lattice-isomorphic to elementary abelian groups.
Added for clarity:
answered Jan 28 at 17:51
the_foxthe_fox
2,90031538
$endgroup$
add a comment |
$begingroup$
Here is the proof in the case of finite abelian groups $G_1, G_2$ like above.
Lemma 1
Let $G^{(n)}$ be the number of elements in $G$ of order $n$. $G^{(n)}$ is uniquely determined by the number of cyclic subgroups of $G$ of order $n$.
Proof
Every element of order $n$ is an element of exactly one cyclic subgroup of $G$ of order $n$. All the cyclic subgroups of order $n$ have $phi(n)$ elements of order $n$.
Lemma 2
Let $p$ be a prime that divides $|G|$ then the numbers $G^{(p)}, G^{(p^2)},...$ uniquely determine the p-Sylow of $G$.
Proof
The p-Sylow, P, of G is of the form $mathbb{Z}_{p^{a_1}} times ... times mathbb{Z}_{p^{a_n}}$. Moreover, let $P^{(leq p^k)}$ be the number of elements of P that have an order less or equal to $p^k$. $$ P^{(leq p^k)}=Pi_{ileq n}{min (p^{a_i}, p^k)}$$
Then we can determine $a_1,...,a_n$.
Then the thesis follows easily.
answered Jan 28 at 23:16
Lucio TanziniLucio Tanzini
351114
$endgroup$
add a comment |
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3 Answers
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3 Answers
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votes
$begingroup$
There are two pairs of examples of order $16$. These are the smallest examples. One of these two pairs is $C_4times C_4$ and $C_4rtimes C_4$. For both of these, the complete list of proper subgroups is:
1 trivial subgroup
3 subgroups isomorphic to $C_2$
6 subgroups isomorphic to $C_4$
1 subgroup isomorphic to $C_2times C_2$
3 subgroups isomorphic to $C_4times C_2$
(See https://groupprops.subwiki.org/wiki/Nontrivial_semidirect_product_of_Z4_and_Z4#Subgroups for the subgroups of $C_4rtimes C_4$.)
Another easy pair of examples is $C_9times C_3$ and $C_9rtimes C_3$.
(It is definitely true for finite abelian groups though, this is an easy consequence of their classification.)
answered Jan 28 at 22:07
verretverret
3,3171923
$endgroup$
add a comment |
$begingroup$
There are two pairs of examples of order $16$. These are the smallest examples. One of these two pairs is $C_4times C_4$ and $C_4rtimes C_4$. For both of these, the complete list of proper subgroups is:
1 trivial subgroup
3 subgroups isomorphic to $C_2$
6 subgroups isomorphic to $C_4$
1 subgroup isomorphic to $C_2times C_2$
3 subgroups isomorphic to $C_4times C_2$
(See https://groupprops.subwiki.org/wiki/Nontrivial_semidirect_product_of_Z4_and_Z4#Subgroups for the subgroups of $C_4rtimes C_4$.)
Another easy pair of examples is $C_9times C_3$ and $C_9rtimes C_3$.
(It is definitely true for finite abelian groups though, this is an easy consequence of their classification.)
answered Jan 28 at 22:07
verretverret
3,3171923
$endgroup$
add a comment |
$begingroup$
There are two pairs of examples of order $16$. These are the smallest examples. One of these two pairs is $C_4times C_4$ and $C_4rtimes C_4$. For both of these, the complete list of proper subgroups is:
1 trivial subgroup
3 subgroups isomorphic to $C_2$
6 subgroups isomorphic to $C_4$
1 subgroup isomorphic to $C_2times C_2$
3 subgroups isomorphic to $C_4times C_2$
(See https://groupprops.subwiki.org/wiki/Nontrivial_semidirect_product_of_Z4_and_Z4#Subgroups for the subgroups of $C_4rtimes C_4$.)
Another easy pair of examples is $C_9times C_3$ and $C_9rtimes C_3$.
(It is definitely true for finite abelian groups though, this is an easy consequence of their classification.)
answered Jan 28 at 22:07
verretverret
3,3171923
$endgroup$
There are two pairs of examples of order $16$. These are the smallest examples. One of these two pairs is $C_4times C_4$ and $C_4rtimes C_4$. For both of these, the complete list of proper subgroups is:
1 trivial subgroup
3 subgroups isomorphic to $C_2$
6 subgroups isomorphic to $C_4$
1 subgroup isomorphic to $C_2times C_2$
3 subgroups isomorphic to $C_4times C_2$
(See https://groupprops.subwiki.org/wiki/Nontrivial_semidirect_product_of_Z4_and_Z4#Subgroups for the subgroups of $C_4rtimes C_4$.)
Another easy pair of examples is $C_9times C_3$ and $C_9rtimes C_3$.
(It is definitely true for finite abelian groups though, this is an easy consequence of their classification.)
answered Jan 28 at 22:07
verretverret
3,3171923
edited Jan 28 at 22:13
answered Jan 28 at 22:07
verretverret
3,3171923
answered Jan 28 at 22:07
verretverret
3,3171923
answered Jan 28 at 22:07
verretverret
3,3171923
3,3171923
add a comment |
add a comment |
$begingroup$
Certainly not always. I'd be surprised if there is a concrete set of conditions which is both necessary and sufficient to conclude isomorphism between the two groups. (My answer refers to finite groups only.)
There are groups which are called $P$-groups in Schmidt's book "Subgroup Lattices of Groups" (not be confused with $p$-groups) and which are lattice-isomorphic to elementary abelian groups.
Added for clarity:
answered Jan 28 at 17:51
the_foxthe_fox
2,90031538
$endgroup$
add a comment |
$begingroup$
Certainly not always. I'd be surprised if there is a concrete set of conditions which is both necessary and sufficient to conclude isomorphism between the two groups. (My answer refers to finite groups only.)
There are groups which are called $P$-groups in Schmidt's book "Subgroup Lattices of Groups" (not be confused with $p$-groups) and which are lattice-isomorphic to elementary abelian groups.
Added for clarity:
answered Jan 28 at 17:51
the_foxthe_fox
2,90031538
$endgroup$
add a comment |
$begingroup$
Certainly not always. I'd be surprised if there is a concrete set of conditions which is both necessary and sufficient to conclude isomorphism between the two groups. (My answer refers to finite groups only.)
There are groups which are called $P$-groups in Schmidt's book "Subgroup Lattices of Groups" (not be confused with $p$-groups) and which are lattice-isomorphic to elementary abelian groups.
Added for clarity:
answered Jan 28 at 17:51
the_foxthe_fox
2,90031538
$endgroup$
Certainly not always. I'd be surprised if there is a concrete set of conditions which is both necessary and sufficient to conclude isomorphism between the two groups. (My answer refers to finite groups only.)
There are groups which are called $P$-groups in Schmidt's book "Subgroup Lattices of Groups" (not be confused with $p$-groups) and which are lattice-isomorphic to elementary abelian groups.
Added for clarity:
answered Jan 28 at 17:51
the_foxthe_fox
2,90031538
edited Jan 28 at 18:02
answered Jan 28 at 17:51
the_foxthe_fox
2,90031538
answered Jan 28 at 17:51
the_foxthe_fox
2,90031538
answered Jan 28 at 17:51
the_foxthe_fox
2,90031538
2,90031538
add a comment |
add a comment |
$begingroup$
Here is the proof in the case of finite abelian groups $G_1, G_2$ like above.
Lemma 1
Let $G^{(n)}$ be the number of elements in $G$ of order $n$. $G^{(n)}$ is uniquely determined by the number of cyclic subgroups of $G$ of order $n$.
Proof
Every element of order $n$ is an element of exactly one cyclic subgroup of $G$ of order $n$. All the cyclic subgroups of order $n$ have $phi(n)$ elements of order $n$.
Lemma 2
Let $p$ be a prime that divides $|G|$ then the numbers $G^{(p)}, G^{(p^2)},...$ uniquely determine the p-Sylow of $G$.
Proof
The p-Sylow, P, of G is of the form $mathbb{Z}_{p^{a_1}} times ... times mathbb{Z}_{p^{a_n}}$. Moreover, let $P^{(leq p^k)}$ be the number of elements of P that have an order less or equal to $p^k$. $$ P^{(leq p^k)}=Pi_{ileq n}{min (p^{a_i}, p^k)}$$
Then we can determine $a_1,...,a_n$.
Then the thesis follows easily.
answered Jan 28 at 23:16
Lucio TanziniLucio Tanzini
351114
$endgroup$
add a comment |
$begingroup$
Here is the proof in the case of finite abelian groups $G_1, G_2$ like above.
Lemma 1
Let $G^{(n)}$ be the number of elements in $G$ of order $n$. $G^{(n)}$ is uniquely determined by the number of cyclic subgroups of $G$ of order $n$.
Proof
Every element of order $n$ is an element of exactly one cyclic subgroup of $G$ of order $n$. All the cyclic subgroups of order $n$ have $phi(n)$ elements of order $n$.
Lemma 2
Let $p$ be a prime that divides $|G|$ then the numbers $G^{(p)}, G^{(p^2)},...$ uniquely determine the p-Sylow of $G$.
Proof
The p-Sylow, P, of G is of the form $mathbb{Z}_{p^{a_1}} times ... times mathbb{Z}_{p^{a_n}}$. Moreover, let $P^{(leq p^k)}$ be the number of elements of P that have an order less or equal to $p^k$. $$ P^{(leq p^k)}=Pi_{ileq n}{min (p^{a_i}, p^k)}$$
Then we can determine $a_1,...,a_n$.
Then the thesis follows easily.
answered Jan 28 at 23:16
Lucio TanziniLucio Tanzini
351114
$endgroup$
add a comment |
$begingroup$
Here is the proof in the case of finite abelian groups $G_1, G_2$ like above.
Lemma 1
Let $G^{(n)}$ be the number of elements in $G$ of order $n$. $G^{(n)}$ is uniquely determined by the number of cyclic subgroups of $G$ of order $n$.
Proof
Every element of order $n$ is an element of exactly one cyclic subgroup of $G$ of order $n$. All the cyclic subgroups of order $n$ have $phi(n)$ elements of order $n$.
Lemma 2
Let $p$ be a prime that divides $|G|$ then the numbers $G^{(p)}, G^{(p^2)},...$ uniquely determine the p-Sylow of $G$.
Proof
The p-Sylow, P, of G is of the form $mathbb{Z}_{p^{a_1}} times ... times mathbb{Z}_{p^{a_n}}$. Moreover, let $P^{(leq p^k)}$ be the number of elements of P that have an order less or equal to $p^k$. $$ P^{(leq p^k)}=Pi_{ileq n}{min (p^{a_i}, p^k)}$$
Then we can determine $a_1,...,a_n$.
Then the thesis follows easily.
answered Jan 28 at 23:16
Lucio TanziniLucio Tanzini
351114
$endgroup$
Here is the proof in the case of finite abelian groups $G_1, G_2$ like above.
Lemma 1
Let $G^{(n)}$ be the number of elements in $G$ of order $n$. $G^{(n)}$ is uniquely determined by the number of cyclic subgroups of $G$ of order $n$.
Proof
Every element of order $n$ is an element of exactly one cyclic subgroup of $G$ of order $n$. All the cyclic subgroups of order $n$ have $phi(n)$ elements of order $n$.
Lemma 2
Let $p$ be a prime that divides $|G|$ then the numbers $G^{(p)}, G^{(p^2)},...$ uniquely determine the p-Sylow of $G$.
Proof
The p-Sylow, P, of G is of the form $mathbb{Z}_{p^{a_1}} times ... times mathbb{Z}_{p^{a_n}}$. Moreover, let $P^{(leq p^k)}$ be the number of elements of P that have an order less or equal to $p^k$. $$ P^{(leq p^k)}=Pi_{ileq n}{min (p^{a_i}, p^k)}$$
Then we can determine $a_1,...,a_n$.
Then the thesis follows easily.
answered Jan 28 at 23:16
Lucio TanziniLucio Tanzini
351114
edited Feb 10 at 15:11
answered Jan 28 at 23:16
Lucio TanziniLucio Tanzini
351114
answered Jan 28 at 23:16
Lucio TanziniLucio Tanzini
351114
answered Jan 28 at 23:16
Lucio TanziniLucio Tanzini
351114
351114
add a comment |
add a comment |
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$begingroup$
$f(A)$ is a subgroup
$endgroup$
– the_fox
Jan 28 at 17:53
1
$begingroup$
If the two groups are not finite we surely can't conclude anything. A counterexample is $mathbb{Z}_2 times mathbb{Z}_4 times mathbb{Z}_4...$ and $mathbb{Z}_4 times mathbb{Z}_4...$
$endgroup$
– Lucio Tanzini
Jan 28 at 17:54
1
$begingroup$
@the_fox Ah, you are right, sets of proper subgroups, this was missing. I am sorry. Still, the question is a bit vague "when can I conclude that $G_1cong G_2$." Certainly not always, but sometimes.
$endgroup$
– Dietrich Burde
Jan 28 at 17:56