Subgroup of an arbitrary group

[fishturtle1]

Homework Statement

Let G be a group. Let H and K be subgroups of G. Prove that if
H $\subseteq$ K, then H is a subgroup of K.

Homework Equations

The Attempt at a Solution

H is a subset of K and H,K are groups.
if x,y, xy $\epsilon$ H, then x,y, xy $\epsilon$ K.
So H is closed under multiplication such that for all x, y, xy $\epsilon$ H,
x, y, xy $\epsilon$ K.

if $x, x^{-1} \epsilon$ H then $x, x^{-1} \epsilon$ K
So H is closed under inverses such that
for all $x^{-1} \epsilon$ H, $x^{-1} \epsilon$ K

H is a subset of K and H is closed under multiplication and closed under inverses so H is also a subgroup of K.

is what I wrote clear?

 

[Orodruin]

You forgot "H contains the unit element", but otherwise fine.
I think you can make this argument much simpler though. You already know that H satisfies all group axioms, since it is a subset of G. Since it is also a subset of K it is a subset of K that satisfies all group axioms and therefore a subgroup of K.

 

[fishturtle1]

You forgot "H contains the unit element", but otherwise fine.
I think you can make this argument much simpler though. You already know that H satisfies all group axioms, since it is a subset of G. Since it is also a subset of K it is a subset of K that satisfies all group axioms and therefore a subgroup of K.

Thanks for showing me the simpler proof. About the unit element, isn't this implied if H is closed under inverses so is mentioning it just good form? Or is there a case where a set is closed under inverses but does not have the unit element?

 

[Orodruin]

Thanks for showing me the simpler proof. About the unit element, isn't this implied if H is closed under inverses so is mentioning it just good form? Or is there a case where a set is closed under inverses but does not have the unit element?

No, you are correct. If the set contains the inverses of all its elements and is closed under the group operation, then the identity must be part of the set. I am just used to ticking off all the group axioms. Also, if you are given a particular subset, it is usually easier to check whether the identity belongs to it than checking if all the inverses do so it might be less work to show that it not a subgroup (if this is the case) just by noting that the identity is not there. Although I have to admit that I omit the associativity property for checking if a group is a subgroup...

 

[fishturtle1]

No, you are correct. If the set contains the inverses of all its elements and is closed under the group operation, then the identity must be part of the set. I am just used to ticking off all the group axioms. Also, if you are given a particular subset, it is usually easier to check whether the identity belongs to it than checking if all the inverses do so it might be less work to show that it not a subgroup (if this is the case) just by noting that the identity is not there. Although I have to admit that I omit the associativity property for checking if a group is a subgroup...

Understood, thank you for the help

 

[WWGD]

A nice result is that a subset S of G is a subgroup of G if for all x,y in S $xy^{-1} \in S$.