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单词 ExampleOfNonpermutableSubgroup
释义

example of non-permutable subgroup


Example 1.

There are groups (even finitely generatedMathworldPlanetmathPlanetmathPlanetmath) with subnormal subgroupsMathworldPlanetmath which arenot permutable.

Proof.

Let D8 be the dihedral groupMathworldPlanetmath of order 8. The classic presentationMathworldPlanetmathPlanetmathPlanetmath is

D8=a,b:a4=b2=1,bab=a-1.

As the group is nilpotentPlanetmathPlanetmath we know every subgroupMathworldPlanetmathPlanetmath is subnormal; however, not every subgroup is permutable. In particular, observe that for two general subgroups H and K of D8, it may be possible that HK is not a subgroup. In this situation we find our counterexample.

bab={1,b,ab,bab}={1,b,ab,a-1}.

Yet

abb={1,ab,b,abb}={1,b,ab,a}.

More generally, in any dihedral group

D2n=a,b:an=b2=1,bab=a-1,

for n>2, then

bababb

and both are subnormal whenever n=2i.∎

However, we do observe the competing observation that the group generatedby b and ab is the same as the groupgenerated by ab and b, namely D8.Indeed in any group with subgroups H and K, H,K=K,H so the condition of permutability is one which must be tested as complexes (sets HK), not as subgroups. This is a consequence of the following general result:

Claim 1.

HK=KH if and only if HK=H,K.

Proof.

We will show that every element in H,K can be written inthe form hk for some hH and kK. To see this first noteevery element in H,K is a word over elements in H and inK. If the word involves only elements in H or only elements in K thenwe are done. Now for inductionMathworldPlanetmath suppose all words of length m in H,K can be expressed in the form hk for some hH and kK.Then given a word of length m+1 we have either hw for hH and wa word of length m, in which case we are done, or kw for some kKand w a word of length m. Then by induction w=hk form some hH andkK. Hence kw=khk. Then khKH=HK so there exists hH and k′′K such that kh=hk′′. Thus kw=hk′′k=h(k′′k) is of thedesired format. Hence HKH,KHK soHK=H,K.

For the converseMathworldPlanetmath suppose HK=H,K. Then KHHK.This means for every kK and hH there exists hH andkK such that k-1h-1=hk. Thus

hk=(k-1h-1)-1=(hk)-1=(k)-1(h)-1KH.

Thus HKKH and KH=HK.∎

This helps illustrate how permutability is such a useful condition in the study of subgroup lattices (one of Ore’s main research interests). For these are the subgroups whose complexes are also subgroups. Thus we can relate the order of H,K to the order of H and K and many other combinatorial relationsMathworldPlanetmath.

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