Prove for $H \subset G$ a subgroup of $G$, the group operation on $G$ restricts to an operation on $H$ making it into a group

abstract-algebragroup-theory

I'm working on a problem and I'm having a bit of trouble getting started. As the title says the problem statement is asking me to prove that

for $H \subset G$ a subgroup of $G$, the group operation on $G$ restricts to an operation on $H$ making it into a group

Now I'm having a bit of trouble understanding exactly what this is asking of me. Particularly I'm getting hung up on the wording of "group operation on $G$ restricts to an operation on $H$"

  • What I imagine 'restricts' in this context means is simply whatever operation that makes $G$ into a group, we now consider it for the subset $H \subset G$

My intuition is that this question is asking us to, given the definition of a subgroup we've been working with:

A nonempty subset $H$ of $G$ such that $H$

  • is closed under the operation $\ast$ : $h \ast g \in H, \forall h,g \in H$
  • is closed under inverses: $\forall g \in H: g^{-1} \in H$

Show that these properties fulfill the properties of a group

A nonempty set $G$ with an operation $\ast$ such that

  • $\ast$ is associative
  • There exists an identity element $e \in G$ such that $e \ast g = g \ast e = g$
  • For all $g \in G$ there exists an inverse element $g^{-1} \in G$ such that $g^{-1} \ast g = g \ast g^{-1} = e$

Is that correct?

If so, would the existence of inverse elements just be a given? Since their existence is just a defining property of a subgroup. Then, with that, the existence of the identity element follows since there can be no inverses without an identity? But then for associativity of $\ast$ im not sure how exactly to show that, unless when we restrict $\ast$ to $H$ we say since $\ast$ is associative on $G$ then it is on $H$ as well. Or, perhaps we say that given $h,g,k \in H$ we have
$$(h \ast g) \ast k = h \ast g \ast k = h \ast (g \ast k)$$
Since because $H$ is closed under $\ast$ we have $(h \ast g),(g \ast k) \in H$

Any help and clarification is greatly appreciated.

Best Answer

It's asking you to show that, for any group $(G,\circ)$ with operation

$$\begin{align} \circ: G\times G&\to G,\\ (x,y)&\mapsto x\circ y, \end{align}$$

any subgroup $H$ of $G$ is itself a group under the operation

$$\begin{align} \circ|_H: H\times H&\to H,\\ (h,k)&\mapsto h\circ k. \end{align}$$

Here $\circ|_H$ is the restriction of $\circ$ to $H$.

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