I am working through Fraleigh's text in abstract algebra.
What things must I show to prove that a subset H of a group G is a subgroup under the group operation?
I thought I had to show:
1) The identity from the group is the identity for the subgroup and is in the subgroup
2) The group is closed under inversion (group operation with inverse element) and that the inverse for the group is the inverse for the subgroup
3) The group is closed under the group product
If I am wrong, please correct me with the proper approach.
Also, what is the easiest way to show that the identity / inverse from the group are the same as in the subgroup? This seems trivial enough as to be difficult.
Best Answer
That's correct, those are the things you need. But there are some shortcuts - regarding the first one, note that no two elements of a group behave like the identity. What I mean is that if $h \in G$ isn't an identity element and $xh = h$, then $x$ is the identity element. That means that in order for $H$ to have an identity element, it has to include the original identity element of $G$ - so all you need to do is check whether that's in $H$.
Likewise, nothing has two inverses - so if $h$ has an inverse in $H$, that's automatically the same inverse as in $G$. So all you really need to do is this:
1) Check whether $e \in H$, where $e$ is the identity element of $G$.
2) Check whether, for each $h \in H$, $h^{-1}$ is also in $H$, where $h^{-1}$ is the inverse of $h$ in $G$.
3) Check whether, for each $h_1,h_2 \in H$, $h_1h_2$ is also in $H$.
If the answer to all three is "yes", then $H$ is a subgroup.