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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.

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What is your definition of a subgroup? –  Joe Johnson 126 Apr 9 '12 at 19:41
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\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
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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

1 Answer 1

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.

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