Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Problem 1:

Prove that a subset $H$ of $G$ is a subgroup if and only if it satisfies the following conditions.

  1. The identity $e$ of $G$ is in $H$.
  2. If $h_1, h_3 \in H$, then $h_1h_2 \in H$.
  3. If $h \in H$, then $h^{-1} \in H$.

Problem 2:

Let $H$ be a subset of a group $G$. Then $H$ is a subgroup of $G$ if and only if $H \neq \emptyset$, and whenever $g, h \in H$ then $gh^{-1}$ is in $H$.

I'm not really sure where to start on these problems - I am just starting to learn abstract algebra on my own.

share|cite|improve this question
What is your definition of a subgroup? – Joe Johnson 126 Apr 9 '12 at 19:41
\emptyset will produce $\emptyset$; \varnothing will produce $\varnothing$ – Arturo Magidin Apr 9 '12 at 19:43
A subgroup $H$ of a group $G$ is a subset $H$ of $G$ such that when the group operation of $G$ is restricted to $H$, $H$ is a group in its own right. – justwatching Apr 9 '12 at 19:44
If the group operation of $G$ restricts to $H$ in a way that $H$ becomes a group, what does that tell you about how to multiply $h_1$ and $h_2$, where $h_1$ and $h_2$ are elements of $H$? Is it possible for $h_1 h_2$ to be in $G$ but not in $H$ if $H$ is a subgroup? If so, can you produce an example? If not, what about the definition prevents that from happening? – Michael Joyce Apr 9 '12 at 19:48
up vote 1 down vote accepted

I assume that your definition of "subgroup" is "a subset which is a group under the operation of $G$."

For problem 1, you just need to show that the conditions imply that $H$ satisfies the axioms of being a group; you will need part 2 to show that multiplication restricts to an operation on $H$; then use part 1 to get that $H$ has an identity; and part 3 gives you the inverses. The converse (that if it is a subgroup then it satisfies these conditions) should be easy.

For problem 2, showing that if $H$ is a subgroup then these two conditions hold should be easy. To show that the conditions imply $H$ is a subgroup, try to show that it satisfies the three conditions in problem 1. Since $H$ is nonempty, there is an element $x\in H$. Now apply the condition with $g=h=x$ (note that we do not require $g$ and $h$ to be distinct!) to conclude that $H$ contains the identity. Then take $h\in H$, and set $g=e$ to conclude that if $h\in H$ then $h^{-1}\in H$. And finally, given $g,h\in H$, use $g$ and $h^{-1}$ to show that $H$ is closed under products.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.