Why need second condition to be a subgroup?

Question

Let H be a non-empty subset of a group G. Then H is a subgroup of G if and only if
1) $a, b \in H$ implies that $ab \in H$, and
2) $a \in H$ implies $a^{−1} \in H$.

I have confusion in the second point as closure is already implied by (1). Then what difference is introduced by (2), by stating that the inverse also is in the set formed by the action table of group.

Answer 1

Consider $G=(\mathbb Z,+)$ and $H=\mathbb N$. It satisfies the first condition, but not the second.

The point is that, unless $G$ is finite, you cannot get inverses or identity by only using closure.

A nice exercise is to prove that for finite groups, the first condition is actually enough.

Answer 2

Because otherwise the subset wouldn't necessarily be a group. For instance, it would allow $\Bbb N\subseteq \Bbb Z$ to be a subgroup, even though it isn't a group. That is clearly unwanted. A sub-something ought to actually be that something, right? A subset should be a set, a subgroup should be a group, a sub-vectorspace (often just called subspace in context) should be a vector space, and so on.

Answer 3

Take the subset $\mathbb{N}$ of the additive group $\mathbb{Z}$. Then by $(1)$ alone, it would be a subgroup. But this is not true, since the inverses of $a$, namely $-a$ are not contained.

Answer 4

Even with closure, we may have $a \in H$ but $a^{-1} \notin H$ since $1)$ is not an if and only if statement. In this case, we cannot call $H$ a subgroup because there is an element $a$ in $H$ such that $a^{-1}$ is not in $H$ and this is against the definition of group.

Answer 5

We want to define a subgroup such that the subgroup is itself a group. Since a group requires inverses, we need inverses in the subgroup as well.

Closure does not imply the existence of inverses. For example, consider the subset $\mathbb{N}$ (the natural numbers, including zero) of the group of integers $\mathbb{Z}$ under addition. $\mathbb{N}$ is closed under addition, but does not contain inverses.

Answer 6

$(2)$ is necessary because otherwise $\mathbb{N}$ would be a subgroup of $\mathbb{Z}$, for instance. You can, however, prove everything with only two conditions:

1. $H\neq \emptyset$
2. $h,g \in H \Longrightarrow hg^{-1} \in H$

Can you see why this second condition is more powerful than just closure under multiplication?

Answer 7

The second condition is not implied by the first. For example the non negative integers as a subset of the additive group of integers satisfy the first property, but not the second. However, both statements are equivalent to: $\forall a,b\in H : a\cdot b^{-1}\in H$