= Subgroup
{title2=$H\le G$}
{wiki}
A <subgroup> $H\le G$ is a <subset> of a <group> $G$ that is itself a <group> under the same <group operation>. It must contain the <identity element> of $G$ and be closed under products and <inverse elements>. <Associativity> is inherited from $G$, so it needs no new proof. The whole <group> and the <trivial group> $\{e\}$ are always <subgroups>. A proper <subgroup> satisfies $H\ne G$.
A useful single-condition test is
$$
H\ne\varnothing\quad\text{and}\quad
ab^{-1}\in H\ \text{for all }a,b\in H.
$$
To see why it works, choose $a\in H$; then $aa^{-1}=e\in H$. Taking $a=e$ gives every <inverse element> $b^{-1}\in H$. Applying the test to $a$ and $b^{-1}$ then gives $ab\in H$. Conversely, every <subgroup> satisfies the test by closure under products and <inverse elements>. This also makes the additive version memorable: a nonempty <subset> of an <additive group> is a <subgroup> exactly when it is closed under differences. See https://ocw.mit.edu/courses/18-703-modern-algebra-spring-2013/ed9964d3df50737e155f0d021af3eb49_MIT18_703S13_pra_l_2.pdf[McKernan's MIT lecture 2, Proposition 2.3, for the equivalent closure criterion].
For the <square> <dihedral group> $D_4$ above, $R=\{e,r,r^2,r^3\}$ is a <cyclic subgroup>: products add exponents modulo four, and $(r^k)^{-1}=r^{-k}$. The smaller collections $\{e,r^2\}$ and $\{e,s\}$ are <subgroups> of <group order> two. But $\{e,r\}$ is not a <subgroup>, since $r\cdot r=r^2$ lies outside it. A plausible-looking list of symmetries therefore need not constitute a complete collection of reversible operations.
Inside $(\mathbb Z,+)$, the <even numbers> $2\mathbb Z$ form a <subgroup>: $2m-2n=2(m-n)$ is an <even number>. Its two <cosets> consist of the <even numbers> and the <odd numbers>, so its <index of a subgroup> is two. The parity <group homomorphism> $k\mapsto k\bmod2$ has $2\mathbb Z$ as its <kernel of a group homomorphism>. A <subgroup> can thus identify exactly which information a composition-preserving map forgets.
<Subgroups> also describe residual symmetry. In the <group action> of $D_4$ on the four vertices, the <stabilizer subgroup> of $(1,1)$ is $\{e,sr\}$: the diagonal <reflection> $sr$ fixes that vertex, and the other six motions move it. Requiring a feature to stay fixed still leaves a <group>, because both composition and reversal preserve that requirement. This is why <subgroups> are useful for studying a <group action> one feature at a time.
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