Do there exist finite non-cyclic groups $H$ and $K$, satisfying the specific condition?

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Let’s define $sigma(G)$ as the sum of orders of all normal subgroups of a finite group $G$. Do there exist two finite groups $H$ and $K$ such, that $sigma(H) = |H| + |K| = sigma(K)$ and $H$ is non-cyclic?



Why $H$ is required to be non-cyclic? A pair of cyclic groups $H$ and $K$ satisfies that condition iff $|H|$ and $|K|$ form an amicable pair. And it would be interesting to know, what happens if at least one of those groups is non-cyclic.







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  • Shouldn't there be some relation between $H, K$ and $G$?
    – Guido A.
    Aug 16 at 11:02






  • 1




    @GuidoA., $G$ is just a symbol for an arbitrary finite group in the definition of a function $sigma(G)$ from finite groups to natural numbers. The only groups that are "present" in this question are $H$ and $K$.
    – Yanior Weg
    Aug 16 at 11:11














up vote
5
down vote

favorite
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Let’s define $sigma(G)$ as the sum of orders of all normal subgroups of a finite group $G$. Do there exist two finite groups $H$ and $K$ such, that $sigma(H) = |H| + |K| = sigma(K)$ and $H$ is non-cyclic?



Why $H$ is required to be non-cyclic? A pair of cyclic groups $H$ and $K$ satisfies that condition iff $|H|$ and $|K|$ form an amicable pair. And it would be interesting to know, what happens if at least one of those groups is non-cyclic.







share|cite|improve this question






















  • Shouldn't there be some relation between $H, K$ and $G$?
    – Guido A.
    Aug 16 at 11:02






  • 1




    @GuidoA., $G$ is just a symbol for an arbitrary finite group in the definition of a function $sigma(G)$ from finite groups to natural numbers. The only groups that are "present" in this question are $H$ and $K$.
    – Yanior Weg
    Aug 16 at 11:11












up vote
5
down vote

favorite
1









up vote
5
down vote

favorite
1






1





Let’s define $sigma(G)$ as the sum of orders of all normal subgroups of a finite group $G$. Do there exist two finite groups $H$ and $K$ such, that $sigma(H) = |H| + |K| = sigma(K)$ and $H$ is non-cyclic?



Why $H$ is required to be non-cyclic? A pair of cyclic groups $H$ and $K$ satisfies that condition iff $|H|$ and $|K|$ form an amicable pair. And it would be interesting to know, what happens if at least one of those groups is non-cyclic.







share|cite|improve this question














Let’s define $sigma(G)$ as the sum of orders of all normal subgroups of a finite group $G$. Do there exist two finite groups $H$ and $K$ such, that $sigma(H) = |H| + |K| = sigma(K)$ and $H$ is non-cyclic?



Why $H$ is required to be non-cyclic? A pair of cyclic groups $H$ and $K$ satisfies that condition iff $|H|$ and $|K|$ form an amicable pair. And it would be interesting to know, what happens if at least one of those groups is non-cyclic.









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edited Aug 16 at 11:15

























asked Aug 16 at 10:58









Yanior Weg

1,0441730




1,0441730











  • Shouldn't there be some relation between $H, K$ and $G$?
    – Guido A.
    Aug 16 at 11:02






  • 1




    @GuidoA., $G$ is just a symbol for an arbitrary finite group in the definition of a function $sigma(G)$ from finite groups to natural numbers. The only groups that are "present" in this question are $H$ and $K$.
    – Yanior Weg
    Aug 16 at 11:11
















  • Shouldn't there be some relation between $H, K$ and $G$?
    – Guido A.
    Aug 16 at 11:02






  • 1




    @GuidoA., $G$ is just a symbol for an arbitrary finite group in the definition of a function $sigma(G)$ from finite groups to natural numbers. The only groups that are "present" in this question are $H$ and $K$.
    – Yanior Weg
    Aug 16 at 11:11















Shouldn't there be some relation between $H, K$ and $G$?
– Guido A.
Aug 16 at 11:02




Shouldn't there be some relation between $H, K$ and $G$?
– Guido A.
Aug 16 at 11:02




1




1




@GuidoA., $G$ is just a symbol for an arbitrary finite group in the definition of a function $sigma(G)$ from finite groups to natural numbers. The only groups that are "present" in this question are $H$ and $K$.
– Yanior Weg
Aug 16 at 11:11




@GuidoA., $G$ is just a symbol for an arbitrary finite group in the definition of a function $sigma(G)$ from finite groups to natural numbers. The only groups that are "present" in this question are $H$ and $K$.
– Yanior Weg
Aug 16 at 11:11










1 Answer
1






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oldest

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up vote
6
down vote



accepted










A quick MAGMA code I wrote finds the following example with $$H=D_10 quad , K=D_19$$ Indeed, in both cases
$sigma(K)=1+19+38=58$
$sigma(H)=1+2+5+10+10+10+20= 58$

and $|H|+|K|=20+38 =58$.



Here is the very naive and probably bugged (the $i^j$ indexing definitely doesn't work, $2^4 = 4^2$ so I am missing some examples, and it checks everything twice which is not efficient) code if you are interested. I checked for examples up to $60$.



N:=60;
A:=AssociativeArray();
A[1]:=1;
for i in [2..N] do
w:=NumberOfSmallGroups(i);
for j in [1..w] do
G:=SmallGroup(i,j);
S:=NormalSubgroups(G);
h:=0;
for k in [1..#S] do
h:=h+Order(S[k]`subgroup);
end for;
A[i^j]:=h;
end for;
end for;
for i in [2..N] do
w:=NumberOfSmallGroups(i);
for j in [1..w] do
for a in [2..N] do
y:=NumberOfSmallGroups(a);
for b in [1..y] do
if A[i^j] eq i+a then
if A[a^b] eq i+a then print "(",i,j,")","(",a,b,")", A[i^j], A[a^b];
end if; end if; end for; end for; end for; end for;


And the output



( 12 1 ) ( 12 1 ) 24 24
( 20 4 ) ( 38 1 ) 58 58
( 24 1 ) ( 28 1 ) 52 52
( 28 1 ) ( 24 1 ) 52 52
( 28 2 ) ( 28 2 ) 56 56
( 30 1 ) ( 30 1 ) 60 60
( 38 1 ) ( 20 4 ) 58 58
( 56 1 ) ( 56 1 ) 112 112


Note that it also contains other examples that might be interesting of "self-amicable groups", such as SmallGroup(30,1) $= C_5 times S_3$.



By self-amicable I mean that they mimic the behaviour of $28$ which forms an amicable pair with itself (realised in group theory by $C_28$).



Edit: found this paper https://arxiv.org/pdf/math/0104012.pdf

Edit 2: found this paper as well http://cage.ugent.be/~tdemedts/preprints/leinster.pdf






share|cite|improve this answer


















  • 1




    Your count for $C_38$ is not right. There is also a subgroup of order $2$.
    – Tobias Kildetoft
    Aug 16 at 11:40






  • 1




    Uh oh, you're right. SmallGroup(38,1) is apparently $D_19$ and not $C_38$ (!!). Thank you
    – AnalysisStudent0414
    Aug 16 at 11:43











  • Ahh, that makes the numbers fit, yeah.
    – Tobias Kildetoft
    Aug 16 at 11:45










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
6
down vote



accepted










A quick MAGMA code I wrote finds the following example with $$H=D_10 quad , K=D_19$$ Indeed, in both cases
$sigma(K)=1+19+38=58$
$sigma(H)=1+2+5+10+10+10+20= 58$

and $|H|+|K|=20+38 =58$.



Here is the very naive and probably bugged (the $i^j$ indexing definitely doesn't work, $2^4 = 4^2$ so I am missing some examples, and it checks everything twice which is not efficient) code if you are interested. I checked for examples up to $60$.



N:=60;
A:=AssociativeArray();
A[1]:=1;
for i in [2..N] do
w:=NumberOfSmallGroups(i);
for j in [1..w] do
G:=SmallGroup(i,j);
S:=NormalSubgroups(G);
h:=0;
for k in [1..#S] do
h:=h+Order(S[k]`subgroup);
end for;
A[i^j]:=h;
end for;
end for;
for i in [2..N] do
w:=NumberOfSmallGroups(i);
for j in [1..w] do
for a in [2..N] do
y:=NumberOfSmallGroups(a);
for b in [1..y] do
if A[i^j] eq i+a then
if A[a^b] eq i+a then print "(",i,j,")","(",a,b,")", A[i^j], A[a^b];
end if; end if; end for; end for; end for; end for;


And the output



( 12 1 ) ( 12 1 ) 24 24
( 20 4 ) ( 38 1 ) 58 58
( 24 1 ) ( 28 1 ) 52 52
( 28 1 ) ( 24 1 ) 52 52
( 28 2 ) ( 28 2 ) 56 56
( 30 1 ) ( 30 1 ) 60 60
( 38 1 ) ( 20 4 ) 58 58
( 56 1 ) ( 56 1 ) 112 112


Note that it also contains other examples that might be interesting of "self-amicable groups", such as SmallGroup(30,1) $= C_5 times S_3$.



By self-amicable I mean that they mimic the behaviour of $28$ which forms an amicable pair with itself (realised in group theory by $C_28$).



Edit: found this paper https://arxiv.org/pdf/math/0104012.pdf

Edit 2: found this paper as well http://cage.ugent.be/~tdemedts/preprints/leinster.pdf






share|cite|improve this answer


















  • 1




    Your count for $C_38$ is not right. There is also a subgroup of order $2$.
    – Tobias Kildetoft
    Aug 16 at 11:40






  • 1




    Uh oh, you're right. SmallGroup(38,1) is apparently $D_19$ and not $C_38$ (!!). Thank you
    – AnalysisStudent0414
    Aug 16 at 11:43











  • Ahh, that makes the numbers fit, yeah.
    – Tobias Kildetoft
    Aug 16 at 11:45














up vote
6
down vote



accepted










A quick MAGMA code I wrote finds the following example with $$H=D_10 quad , K=D_19$$ Indeed, in both cases
$sigma(K)=1+19+38=58$
$sigma(H)=1+2+5+10+10+10+20= 58$

and $|H|+|K|=20+38 =58$.



Here is the very naive and probably bugged (the $i^j$ indexing definitely doesn't work, $2^4 = 4^2$ so I am missing some examples, and it checks everything twice which is not efficient) code if you are interested. I checked for examples up to $60$.



N:=60;
A:=AssociativeArray();
A[1]:=1;
for i in [2..N] do
w:=NumberOfSmallGroups(i);
for j in [1..w] do
G:=SmallGroup(i,j);
S:=NormalSubgroups(G);
h:=0;
for k in [1..#S] do
h:=h+Order(S[k]`subgroup);
end for;
A[i^j]:=h;
end for;
end for;
for i in [2..N] do
w:=NumberOfSmallGroups(i);
for j in [1..w] do
for a in [2..N] do
y:=NumberOfSmallGroups(a);
for b in [1..y] do
if A[i^j] eq i+a then
if A[a^b] eq i+a then print "(",i,j,")","(",a,b,")", A[i^j], A[a^b];
end if; end if; end for; end for; end for; end for;


And the output



( 12 1 ) ( 12 1 ) 24 24
( 20 4 ) ( 38 1 ) 58 58
( 24 1 ) ( 28 1 ) 52 52
( 28 1 ) ( 24 1 ) 52 52
( 28 2 ) ( 28 2 ) 56 56
( 30 1 ) ( 30 1 ) 60 60
( 38 1 ) ( 20 4 ) 58 58
( 56 1 ) ( 56 1 ) 112 112


Note that it also contains other examples that might be interesting of "self-amicable groups", such as SmallGroup(30,1) $= C_5 times S_3$.



By self-amicable I mean that they mimic the behaviour of $28$ which forms an amicable pair with itself (realised in group theory by $C_28$).



Edit: found this paper https://arxiv.org/pdf/math/0104012.pdf

Edit 2: found this paper as well http://cage.ugent.be/~tdemedts/preprints/leinster.pdf






share|cite|improve this answer


















  • 1




    Your count for $C_38$ is not right. There is also a subgroup of order $2$.
    – Tobias Kildetoft
    Aug 16 at 11:40






  • 1




    Uh oh, you're right. SmallGroup(38,1) is apparently $D_19$ and not $C_38$ (!!). Thank you
    – AnalysisStudent0414
    Aug 16 at 11:43











  • Ahh, that makes the numbers fit, yeah.
    – Tobias Kildetoft
    Aug 16 at 11:45












up vote
6
down vote



accepted







up vote
6
down vote



accepted






A quick MAGMA code I wrote finds the following example with $$H=D_10 quad , K=D_19$$ Indeed, in both cases
$sigma(K)=1+19+38=58$
$sigma(H)=1+2+5+10+10+10+20= 58$

and $|H|+|K|=20+38 =58$.



Here is the very naive and probably bugged (the $i^j$ indexing definitely doesn't work, $2^4 = 4^2$ so I am missing some examples, and it checks everything twice which is not efficient) code if you are interested. I checked for examples up to $60$.



N:=60;
A:=AssociativeArray();
A[1]:=1;
for i in [2..N] do
w:=NumberOfSmallGroups(i);
for j in [1..w] do
G:=SmallGroup(i,j);
S:=NormalSubgroups(G);
h:=0;
for k in [1..#S] do
h:=h+Order(S[k]`subgroup);
end for;
A[i^j]:=h;
end for;
end for;
for i in [2..N] do
w:=NumberOfSmallGroups(i);
for j in [1..w] do
for a in [2..N] do
y:=NumberOfSmallGroups(a);
for b in [1..y] do
if A[i^j] eq i+a then
if A[a^b] eq i+a then print "(",i,j,")","(",a,b,")", A[i^j], A[a^b];
end if; end if; end for; end for; end for; end for;


And the output



( 12 1 ) ( 12 1 ) 24 24
( 20 4 ) ( 38 1 ) 58 58
( 24 1 ) ( 28 1 ) 52 52
( 28 1 ) ( 24 1 ) 52 52
( 28 2 ) ( 28 2 ) 56 56
( 30 1 ) ( 30 1 ) 60 60
( 38 1 ) ( 20 4 ) 58 58
( 56 1 ) ( 56 1 ) 112 112


Note that it also contains other examples that might be interesting of "self-amicable groups", such as SmallGroup(30,1) $= C_5 times S_3$.



By self-amicable I mean that they mimic the behaviour of $28$ which forms an amicable pair with itself (realised in group theory by $C_28$).



Edit: found this paper https://arxiv.org/pdf/math/0104012.pdf

Edit 2: found this paper as well http://cage.ugent.be/~tdemedts/preprints/leinster.pdf






share|cite|improve this answer














A quick MAGMA code I wrote finds the following example with $$H=D_10 quad , K=D_19$$ Indeed, in both cases
$sigma(K)=1+19+38=58$
$sigma(H)=1+2+5+10+10+10+20= 58$

and $|H|+|K|=20+38 =58$.



Here is the very naive and probably bugged (the $i^j$ indexing definitely doesn't work, $2^4 = 4^2$ so I am missing some examples, and it checks everything twice which is not efficient) code if you are interested. I checked for examples up to $60$.



N:=60;
A:=AssociativeArray();
A[1]:=1;
for i in [2..N] do
w:=NumberOfSmallGroups(i);
for j in [1..w] do
G:=SmallGroup(i,j);
S:=NormalSubgroups(G);
h:=0;
for k in [1..#S] do
h:=h+Order(S[k]`subgroup);
end for;
A[i^j]:=h;
end for;
end for;
for i in [2..N] do
w:=NumberOfSmallGroups(i);
for j in [1..w] do
for a in [2..N] do
y:=NumberOfSmallGroups(a);
for b in [1..y] do
if A[i^j] eq i+a then
if A[a^b] eq i+a then print "(",i,j,")","(",a,b,")", A[i^j], A[a^b];
end if; end if; end for; end for; end for; end for;


And the output



( 12 1 ) ( 12 1 ) 24 24
( 20 4 ) ( 38 1 ) 58 58
( 24 1 ) ( 28 1 ) 52 52
( 28 1 ) ( 24 1 ) 52 52
( 28 2 ) ( 28 2 ) 56 56
( 30 1 ) ( 30 1 ) 60 60
( 38 1 ) ( 20 4 ) 58 58
( 56 1 ) ( 56 1 ) 112 112


Note that it also contains other examples that might be interesting of "self-amicable groups", such as SmallGroup(30,1) $= C_5 times S_3$.



By self-amicable I mean that they mimic the behaviour of $28$ which forms an amicable pair with itself (realised in group theory by $C_28$).



Edit: found this paper https://arxiv.org/pdf/math/0104012.pdf

Edit 2: found this paper as well http://cage.ugent.be/~tdemedts/preprints/leinster.pdf







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 17 at 12:56

























answered Aug 16 at 11:36









AnalysisStudent0414

4,281828




4,281828







  • 1




    Your count for $C_38$ is not right. There is also a subgroup of order $2$.
    – Tobias Kildetoft
    Aug 16 at 11:40






  • 1




    Uh oh, you're right. SmallGroup(38,1) is apparently $D_19$ and not $C_38$ (!!). Thank you
    – AnalysisStudent0414
    Aug 16 at 11:43











  • Ahh, that makes the numbers fit, yeah.
    – Tobias Kildetoft
    Aug 16 at 11:45












  • 1




    Your count for $C_38$ is not right. There is also a subgroup of order $2$.
    – Tobias Kildetoft
    Aug 16 at 11:40






  • 1




    Uh oh, you're right. SmallGroup(38,1) is apparently $D_19$ and not $C_38$ (!!). Thank you
    – AnalysisStudent0414
    Aug 16 at 11:43











  • Ahh, that makes the numbers fit, yeah.
    – Tobias Kildetoft
    Aug 16 at 11:45







1




1




Your count for $C_38$ is not right. There is also a subgroup of order $2$.
– Tobias Kildetoft
Aug 16 at 11:40




Your count for $C_38$ is not right. There is also a subgroup of order $2$.
– Tobias Kildetoft
Aug 16 at 11:40




1




1




Uh oh, you're right. SmallGroup(38,1) is apparently $D_19$ and not $C_38$ (!!). Thank you
– AnalysisStudent0414
Aug 16 at 11:43





Uh oh, you're right. SmallGroup(38,1) is apparently $D_19$ and not $C_38$ (!!). Thank you
– AnalysisStudent0414
Aug 16 at 11:43













Ahh, that makes the numbers fit, yeah.
– Tobias Kildetoft
Aug 16 at 11:45




Ahh, that makes the numbers fit, yeah.
– Tobias Kildetoft
Aug 16 at 11:45












 

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