Subgroup ($H\le G$)

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 isTo see why it works, choose $a\in H$; then $a{a}^{-1}=e\in H$. 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 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 $k\mapsto k\mathrm{mod}2$ 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.

A nonempty subset $H$ of a group $G$ is a subgroup if and only if $g{h}^{-1}\in H$ for every $g,h\in H$. Necessity follows from closure and the existence of inverse elements in a subgroup.

Conversely, choose $a\in H$. Setting $g=h=a$ gives the identity element $e=a{a}^{-1}\in H$. Taking $g=e$ then gives $h^{-1}\in H$ for every $h\in H$. Finally, using $h^{-1}$ as the second element gives $g(h^{-1}{)}^{-1}=gh\in H$, proving closure. The inherited multiplication is associative, so these properties establish the group axioms on $H$.

The trivial subgroup contains only the identity element of a group. It is a subgroup of every group and is proper whenever the group has more than one element. Thus a group of prime group order has no proper nontrivial subgroups, rather than having no proper subgroups at all.

A subgroup is finitely generated when finitely many of its elements generate it as a group. Even inside a finitely generated group, subgroups need not be finitely generated. Finite generation of a witness subgroup, rather than of the original intersection, is what makes centralizing HNN extensions usable for a benign subgroup.

For a specified subgroup $H\le G$ and a chosen alphabet for $G$, the membership problem asks whether an input word represents an element of $H$. Solubility of the ordinary word problem for a group does not imply solubility of this problem, even for finitely generated subgroups: a Mihailova subgroup of a product of finite-rank free groups can have undecidable membership.

A proper subgroup with no strictly intermediate subgroup between it and the ambient group. In a transitive action a point stabilizer is maximal exactly when the action is primitive.

Up to conjugation, the proper maximal subgroups of $S_n$ containing a three-cycle are $A_n$; the unequal-part subset stabilizers $S_k\times S_{n-k}$ for $1\le k<n/2$ with $n-k\ge 3$; and the full block stabilizers $S_a\wr S_b$ for $n=ab$, $a\ge 3$, $b\ge 2$. The primitive three-cycle criterion treats the primitive case. A nontrivial block system gives the full permutation wreath product in the imprimitive case. These block stabilizers are maximal because any extra permutation produces a transposition connecting two different blocks; conjugating by the block stabilizer then gives every cross-block transposition, hence the symmetric group. Unequal-part subset stabilizers are maximal by the same connected-transposition argument. Equal-part subset stabilizers are not maximal: they lie in the larger stabilizer that can interchange the two parts.

Two maximal subgroups of a finite soluble group either multiply to the whole group or are conjugate. Reduce modulo a nontrivial normal subgroup contained in either maximum. If no such subgroup exists, both maxima complement a minimal normal elementary abelian subgroup $K$. Their actions on $K$ are faithful and irreducible. A minimal normal subgroup of $G/K$ has characteristic different from that of $K$, since a normal p-subgroup has nonzero fixed points on a characteristic-p module. The intersections of the two maxima with its preimage are conjugate Sylow subgroups. Each maximum is the normalizer of its nontrivial intersection, proving conjugacy.

For a maximal subgroup $M$ of a finite nonabelian simple group, take a minimal nontrivial normal subgroup $N$ of $M$. It is a characteristically simple group. Its normalizer contains $M$ but cannot be the whole simple group, since $1<N\le M<G$. Maximality gives equality.

A Hall subgroup has order coprime to its index. For a set of primes $\pi$, a Hall $\pi$-subgroup has order using only primes of $\pi$ and index using only primes outside $\pi$. Thus its order contains the full prime-power contributions belonging to $\pi$. A Sylow subgroup is the single-prime instance. Hall subgroup existence in soluble groups generalizes Sylow existence to every prime set.

A finite normal subgroup whose order is coprime to its index has a group complement: if $N◃G$ is a Hall subgroup, then $G=NH$ and $N\cap H=1$ for some subgroup $H$. Complement conjugacy holds when either $N$ or $G/N$ is solvable; the existence assertion does not require that extra hypothesis.

For any prime set $\pi$, every finite soluble group has Hall $\pi$-subgroups, any two are conjugate, and every $\pi$-subgroup lies in one. Induction uses an elementary abelian minimal normal subgroup $N$ of characteristic $p$. If $p\in \pi$, the preimage of a quotient Hall subgroup is itself a Hall subgroup. If $p\notin \pi$, coprime splitting over an elementary abelian normal subgroup gives a conjugate class of complements in that preimage. The same complement conjugacy in the preimage of a smaller quotient subgroup gives the embedding assertion.

Every finite soluble group has a Hall subgroup for each prime set. Induct on the group order using an elementary abelian minimal normal subgroup. In the coprime hard case, lift a minimal normal subgroup of the quotient, choose its Sylow subgroup, and apply the Frattini argument. A proper normalizer reduces the order; a normal Sylow subgroup allows induction in its quotient. This proves existence without assuming an independent complement theorem.

The union of two subgroups is a subgroup exactly when one contains the other. Otherwise choose $h\in H\setminus K$ and $k\in K\setminus H$. Their product cannot lie in $H$, since $h^{-1}(hk)=k$ would then lie in $H$, and cannot lie in $K$, since $(hk)k^{-1}=h$ would then lie in $K$. The union therefore fails closure. Unlike the union, the subgroup generated by both subgroups always exists.

The index $[G:H]$ is the number of left cosets of $H$ in $G$.

Let $G$ be finite and let $H$ have index $p$, the smallest prime divisor of $|G|$. The group action on left cosets gives $\Phi :G\to S_p$, with $\ker \Phi \subseteq H$. The image order divides both $|G|$ and $p!$. By the choice of $p$, the only possible nontrivial prime factor common to these orders is $p$, and $p!$ contains only one factor of $p$. Thus the image order is $1$ or $p$. The action is transitive on $p>1$ cosets, so its image is nontrivial and has order $p$. The first isomorphism theorem gives $|\ker \Phi |=|G|/p=|H|$, hence $\ker \Phi =H$. A kernel of a group homomorphism is a normal subgroup, proving the claim. For example, every subgroup of order $25$ in a group of order $75$ is normal.

Every subgroup $H\le G$ of index two is normal: the two left cosets and the two right cosets are both $H$ and its complement, so they agree.

A subgroup $H\le G$ has finite index when the coset space $G/H$ is finite. Intersecting a finite-index subgroup with a subgroup again gives a finite-index subgroup of the latter.

In an infinite group $G$, the complement of a finite union $H_1\cup \cdots \cup H_k$ of finite-index subgroups is either empty or infinite. To prove this, their intersection $L$ has finite subgroup index by the finite intersection of finite-index subgroups bound, so it is infinite: otherwise finitely many finite left cosets would cover $G$. If $g$ lies outside the union, the whole infinite left coset $gL$ does too. Indeed, $g\ell \in H_i$ and $\ell \in L\subseteq H_i$ would imply $g\in H_i$.

For finite-index subgroups $H,K$ of a group $G$, the map of sets of left cosetsis well defined and injective: equality of both components is equivalent to the difference $g^{\prime -1}g$ belonging to $H\cap K$. Hence the displayed subgroup index bound holds, without a normal subgroup hypothesis. By mathematical induction, the index of a subgroup for a finite intersection is at most the product of the individual subgroup indices.