# Subgroup nesting

• MHB
• Jen917

#### Jen917

Let H be a subgroup of G and let L be a subgroup of H. Prove that K is a subgroup of G.

This question seems very redundant to me, isn't anything in a subgroup automatically a subgroup of anything the larger group is a subgroup of. Can some one explain this proof to me?

Let H be a subgroup of G and let L be a subgroup of H. Prove that K is a subgroup of G.

This question seems very redundant to me, isn't anything in a subgroup automatically a subgroup of anything the larger group is a subgroup of. Can some one explain this proof to me?

Hi again Jen917! (Wave)

Can I assume there is a typo, and that K and L are actually the same?

Then admittedly, the proof is pretty straight forward.
It's just that in math we can't assume that it's redundant.

To conclude that one set is a subgroup of another, we have to apply the definition of a subgroup, and verify if all conditions are fulfilled.
This would be an exercise in carefully reading and applying a definition.
Which proof do you have?

Hi again Jen917! (Wave)

Can I assume there is a typo, and that K and L are actually the same?

Then admittedly, the proof is pretty straight forward.
It's just that in math we can't assume that it's redundant.

>>Gotcha, I think I was overthinking it. The easy answer seemed just too easy!

To conclude that one set is a subgroup of another, we have to apply the definition of a subgroup, and verify if all conditions are fulfilled.
This would be an exercise in carefully reading and applying a definition.
Which proof do you have?

This was my idea:
K is a nonempty set and has an identity element, we know this because it is a subgroup of H.
K also contains the inverse of an element following the same logic.
Finally we know K is closed because it is a subgroup of H.
H is a subgroup of G, therefore K has all these characteristics within G and is a subgroup of G.

Is this the along the right idea?

Yep. (Nod)

Nitpick: the sentence about the inverse should be about 'any' element rather than 'an' element.