Do subgrps inherent their group properties

[Bachelier]

because the only thing the definition asks to check is the closure and inverse axioms?

this arose from a problem I was working on. elements with infinite order of an abelian grp G do not necessarly make a subgrp with 0. counter example: Z x Z3
consider (1,1),(-1,1) both with inf. order but their sum (0,2) has order 3.

sorry for the abbrev. I am sending this from my phone

 

[Deveno]

and the answer is:

"sometimes"

for example, the subgroup of an abelian group is always abelian, but the subgroup of a non-abelian group may very well be abelian.

a group may be infinite, and yet have finite subgroups. for example, the non-zero reals under multiplication is an infinite group, but {-1,1} is a finite (and cyclic!) subgroup.

cyclic groups are the nicest. every subgroup of a cyclic group is also cyclic.

most group properties classify groups into "one type of thing" and "non-type of thing". it is possible to have group properties that are lost when you go to a subgroup, and also to have properties that are gained when you go to a subgroup.

for example, not every subgroup of S4 is abelian, because S4 has a subgroup isomorphic to S3, which is clearly not abelian. but...every subgroup of S3 IS abelian.

infinite groups tend to be "more unpredictable" than finite groups. they have lots and lots of subgroups, which can vary widely in character. whereas finite groups, because they are severely restricted by their order, are often easy to "dissect", and the kinds of possible subgroups are likewise restricted. with certain exceptions (like groups of prime order), the smaller a group is, the more likely you are to be able to know for certain which group you have (groups of orders 1,2,3,5 and 7 are all very predictable, and for groups of orders 4 and 6, there's also not much variety. 8 is interesting).

a good group to look at, to get more of a "feel" for what I'm saying, is S4. S4 has a lot of interesting subgroups, but not so many that it's overwhelming to look at them: it has subgroups of orders 12,8,6,4,3,2 and 1. some of these subgroups are abelian, some are not. some are normal (i don't know if you know what that means yet), and some aren't. it's pretty amazing that just considering re-arrangements of 4 things, we can come up with so much structure.

 

[lavinia]

because the only thing the definition asks to check is the closure and inverse axioms?

this arose from a problem I was working on. elements with infinite order of an abelian grp G do not necessarly make a subgrp with 0. counter example: Z x Z3
consider (1,1),(-1,1) both with inf. order but their sum (0,2) has order 3.

sorry for the abbrev. I am sending this from my phone

not generally but there are cases where they do .For instance every subgroup of a free group is free. Every subgroup of a free abelian group is free abelian.

 

[Bachelier]

Thanks both for the answer. Yes D. I know what normal is.

But in this case, why can't we make a subgroup out of infinite elements of Z x Z/3Z ?

 

[Deveno]

well, for one, the identity doesn't have infinite order, right?

and as you demonstrated, we don't even have closure. closure is important.

on the other hand, there IS a subgroup of Z x Z/3Z composed of finite elements, it is called

the torsion subgroup.