# Symmetric Group

Deveno
Science Advisor
I've just thinking about the previous part.

$S_3 \times S_3 = \{ e, (12), (23), (13), (123), (132) \} \times \{ e, (45), (56), (46), (456), (465) \}$

is not a copy of $S_3 \times S_3$ inside $S_6$ because it is not a subgroup of $S_6$ (subgroups of $S_6$ consist of elements of $S_6$).

How can I convert the pairs of elements into elements of $S_6$?

that's a good point.

this is a specific example of a more general theorem:

suppose A,B are subgroups of a finite group G such that:

ab = ba, for all a in A and b in B, and that:

A∩B = {e}.

prove that:

a) AB = {ab: a in A, b in B} is a subgroup of G
b) AxB is isomorphic to AB

note that since G is finite, you only need to show closure to prove (a). to prove (b), it suffices to show that if ab = e, then a = b = e (why?).

that's a good point.

this is a specific example of a more general theorem:

suppose A,B are subgroups of a finite group G such that:

ab = ba, for all a in A and b in B, and that:

A∩B = {e}.

prove that:

a) AB = {ab: a in A, b in B} is a subgroup of G
b) AxB is isomorphic to AB

note that since G is finite, you only need to show closure to prove (a). to prove (b), it suffices to show that if ab = e, then a = b = e (why?).

I remember a 'direct product theorem' from my Groups and Rings course that said if $A,B<G$ satisfy

(i) $A,B \triangleleft G$
(ii) $G=AB$
(iii) $A\cap B = \{ e\}$

then $G \cong A\times B$

Now the 2 subgroups in my case are normal subgroups since they are abelian and certainly $A \cap B = \{e\}$. But you say that it is enough for $AB<G$ rather than $G=AB$.

Deveno
Science Advisor
I remember a 'direct product theorem' from my Groups and Rings course that said if $A,B<G$ satisfy

(i) $A,B \triangleleft G$
(ii) $G=AB$
(iii) $A\cap B = \{ e\}$

then $G \cong A\times B$

Now the 2 subgroups in my case are normal subgroups since they are abelian and certainly $A \cap B = \{e\}$. But you say that it is enough for $AB<G$ rather than $G=AB$.

in the case at hand S3xS3 couldn't possibly be isomorphic to all of S6, the first has only 36 elements, while the latter has 720 (yes, that many).

it is NOT true that "abelian subgroup = normal subgroup".

what IS true is "subgroup of abelian group = normal subgroup".

for example, {e,(1 2)} is an abelian subgroup of S3, but it is NOT normal.

i was careful (at least i HOPE i was) NOT to mention normality at all in my statement. neither A NOR B need be normal subgroups of G in order for AB to be a subgroup. in fact, S3 ISN'T a normal subgroup of S6:

(1 4)(1 2 3)(1 4)-1 = (1 4)(1 2 3)(1 4) = (2 3 4):

1→4→4→1
2→2→3→3
3→3→1→4
4→1→2→2

so the subgroup of S6 that fixes {4,5,6} isn't normal, because i just exhibited a conjugate of (1 2 3) that moves 4.

if G is FINITE (i DID mention this), that the only condition for AB to be a subgroup is closure:

(ab)(a'b') must be in AB.

the trouble is, (ab)(a'b') = a(ba')b' and it's the ba' part that gives us the trouble. we want to get "all the a's on the left, and all the b's on the right".

now, if G was abelian, that would be no trouble at all. but, S6 isn't abelian. but if elements of A and B commute (this is not the same as saying everything in G commutes), then....?

in the case at hand S3xS3 couldn't possibly be isomorphic to all of S6, the first has only 36 elements, while the latter has 720 (yes, that many).

it is NOT true that "abelian subgroup = normal subgroup".

what IS true is "subgroup of abelian group = normal subgroup".

for example, {e,(1 2)} is an abelian subgroup of S3, but it is NOT normal.

i was careful (at least i HOPE i was) NOT to mention normality at all in my statement. neither A NOR B need be normal subgroups of G in order for AB to be a subgroup. in fact, S3 ISN'T a normal subgroup of S6:

(1 4)(1 2 3)(1 4)-1 = (1 4)(1 2 3)(1 4) = (2 3 4):

1→4→4→1
2→2→3→3
3→3→1→4
4→1→2→2

so the subgroup of S6 that fixes {4,5,6} isn't normal, because i just exhibited a conjugate of (1 2 3) that moves 4.

if G is FINITE (i DID mention this), that the only condition for AB to be a subgroup is closure:

(ab)(a'b') must be in AB.

the trouble is, (ab)(a'b') = a(ba')b' and it's the ba' part that gives us the trouble. we want to get "all the a's on the left, and all the b's on the right".

now, if G was abelian, that would be no trouble at all. but, S6 isn't abelian. but if elements of A and B commute (this is not the same as saying everything in G commutes), then....?

$(ab)(a'b') = a(ba')b' = a(a'b)b' = (aa')(bb') \in AB$