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Let $(G,\ast,e)$ be a group, and $(H,\ast,e)$ and $(K,\ast,e)$ be subgroups of $G$. If $K\subseteq H$, is $K \leq H$ ?

With the subgroup test, it seems trivial, but I just want to make sure I'm not making some stupid mistake, as I'm using this for another proof that's causing me to question everything.

Shaun
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Sprinkle
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    Yes, that's true. – Stefan Mesken Aug 10 '16 at 21:16
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    Yeah. The definition of subgroup is that it's a group (check) and a subset (check). It's kind of .... meaningless. To talk of K being a group either all elements also elements of H but not a subgroup. How exactly can a group not be a subgroup except by not being a subset. – fleablood Aug 11 '16 at 01:44

2 Answers2

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Yes, that is true. It follows immediately from one step subgroup test, that is

For K subset of H,

K is a subgroup of H if and only if:

  1. K is not empty;
  2. For every two elements $a,b\in K$ also $a^{-1}b\in K$

Can you check that these two conditions hold?

Salech Alhasov
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Assuming a fixed operation $*$ which is the same for all groups in the context, the statement $K$ is a subgroup of $H$ just means that (i) $K$ is a subset of $H$ AND (ii) $K$ is a group.

This applies in this case since $G, H, K$ are all using the same operation. Since $K$ is a subgroup of $G$, it's a group, and since it's a subset of $H$, by the above fact, it's a subgroup of $H$.

Caleb Stanford
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