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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 subgroups $H,P\le G$, the double coset of $x\in G$ isThe double cosets partition $G$: they are the group orbits of left multiplication by $H$ and right multiplication by $P$. In the group action of $H$ on the left cosets of $P$, the stabilizer subgroup of $xP$ is $H\cap xP{x}^{-1}$. Thus $HxP$ contains $[H:H\cap xP{x}^{-1}]$ left cosets of $P$.

For finite $H,P$, the orbit-stabilizer theorem gives the numerical formulaThe index statement remains meaningful for infinite groups, without dividing infinite cardinalities.

A nonempty subset $B$ of any group $G$ is a left coset of a subgroup if and only if any two left translates $gB,hB$ are equal or disjoint. No finiteness assumption is needed.

For the forward direction, write $B=bH$ with $H$ a subgroup. Then $gB=gbH$, and distinct left cosets of $H$ are disjoint. For the converse, first assume $e\in B$. For every $b\in B$, $b\in B\cap bB$, so $bB=B$. This proves closure under multiplication. Since $e\in B=bB$, there is $c\in B$ with $bc=e$, showing $b^{-1}\in B$. Thus $B$ passes the subgroup test.

For a general nonempty $B$, choose $b\in B$ and set $H=b^{-1}B$. It contains $e$, and all its left translates have the assumed property because $gH=(g{b}^{-1})B$. The previous argument makes $H$ a subgroup, and $B=bH$ is a left coset. This gives a recognition criterion for cosets from their translate intersections alone.

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.

Lagrange's theorem states that for a subgroup $H$ of a finite group $G$,where $[G:H]$ is the index of a subgroup, the number of its left cosets. In particular, $|H|$ divides $|G|$.

The cyclic subgroup $⟨g⟩$ consists of the distinct elements $e,g,\dots ,g^{n-1}$, where $n=o(g)$. Every power reduces to one of these by the Euclidean division, and a repetition among them would contradict minimality of $n$. Thus $|⟨g⟩|=o(g)$. Applying Lagrange's theorem gives

By Lagrange's theorem, the index of a subgroup is $[G:H]=75/25=3$. Thus the set of left cosets has three elements, and the preceding group action yields a group homomorphism $\Phi :G\to S_3$ after labelling them.

Its image is a subgroup of $S_3$, so its order divides $6$. By the first isomorphism theorem, $|\mathrm{im}\Phi |=|G|/|\ker \Phi |$ also divides $75$. Hence its order divides $\gcd (6,75)=3$, giving possibilities $1$ and $3$. The action on cosets is transitive: $g_2{g}_1^{-1}$ sends $g_{1}H$ to $g_{2}H$. It cannot be the trivial action on a three-element set, so the image does not have order $1$. ThereforeIt follows that $|\ker \Phi |=75/3=25$. Part (b)(i) gives $\ker \Phi \le H$, and $|H|=25$; equal finite orders force equality. A kernel of a group homomorphism is a normal subgroup, soThis is an instance of the fact that a subgroup of index the smallest prime divisor is normal.

Define $g\cdot (g_{1}H)=(g{g}_1)H$ on the set of left cosets. It is well-defined: if $g_{1}H=g_{2}H$, then $g_2=g_{1}h$ for some $h\in H$, so $(g{g}_2)H=(g{g}_{1}h)H=(g{g}_1)H$. The identity fixes each coset, and associativity in $G$ givesHence this is a left group action, and part (a) supplies its group homomorphism $\Phi :G\to \mathrm{Sym}(S)$.

An element of the kernel of a group homomorphism $\Phi$ fixes every coset, in particular $H$ itself. Thus $g\in \ker \Phi$ implies $gH=H$, which implies $g\in H$. ThereforeThe subgroup symbol is justified because a kernel of a group homomorphism is a subgroup, and the argument proves containment.

Lagrange's theorem states that if $H$ is a subgroup of a finite group $G$, thenwhere the index of a subgroup $[G:H]$ is the number of distinct left cosets.

Every $g\in G$ lies in $gH$, since $g=ge$. By i, distinct left cosets are disjoint, so these sets form a partition of a set $G$. Each has cardinality $|H|$: the function $h\mapsto gh$ from $H$ to $gH$ is a bijection, with inverse $gh\mapsto h$, by cancellation in a group. Counting the elements in this finite partition of a set gives $|G|=[G:H]|H|$, proving the theorem.

For a subgroup $H$ of $G$, the left coset represented by $g$ isSuppose two left cosets intersect. There are $h_1,h_2\in H$ such that $g_1{h}_1=g_2{h}_2$, soBecause $H$ is a subgroup, $h_2{h}_1^{-1}\in H$. Thus, for every $h\in H$, $g_{1}h=g_2(h_2{h}_1^{-1}h)\in g_{2}H$, proving $g_{1}H\subseteq g_{2}H$. Interchanging the two representatives gives the reverse inclusion. Therefore two left cosets are equal or disjoint.

Choose $b_0\in B$ and put $H=b_0^{-1}B$. Then $e\in H$. For any $g_1,g_2\in G$,so these translates are equal or disjoint by the assumed property of $B$. Applying i shows that $H$ is a subgroup. ThusThis proves that $B$ is a left coset. Together with i, it gives the equal-or-disjoint translates characterize cosets criterion.

The orbit-stabilizer theorem identifies the orbit of a group action of $s$ with the set of left cosets of $G_s$. Explicitly, there is a bijectionHere $G/G_s$ denotes a set of cosets; it is not asserted to be a quotient group. To prove the identification, observe thatThis equivalence proves both well-definedness and injectivity. Every element of the orbit of a group action has the form $g\cdot s$, proving surjectivity.

In particular, for a finite group, the cardinalities satisfyThe last equality follows from Lagrange's theorem applied to the stabilizer subgroup.

Every left coset $kH$ with $k\in K$ is also a left coset of $H$ in $G$. Moreover equality of two such cosets means the same thing in $K$ and in $G$:Thus the map from the set of left cosets of $H$ in $K$ to the set of left cosets of $H$ in $G$, taking $kH$ to itself, is injective. Consequently the index of a subgroup satisfiesThese are sets of cosets; no normal subgroup assumption is needed.

Suppose $H$ were finite, say $|H|=r$. There are at most $n$ left cosets, each with $r$ elements by part (i), and they partition $G$. Hence $G$ would be a finite set with at most $nr$ elements. This contradicts the assumption that $G$ is an infinite group. Therefore $\boxed{H\text{ is infinite}}$.

For every $g\in G$, left multiplication gives a bijectionwhose inverse is $x\mapsto g^{-1}x$. Hence every left coset has the same cardinality as $H$. If $H$ is an infinite set, every left coset is infinite; if $H$ is a finite set, every left coset is finite. These are the only possibilities.

Define a map between sets of left cosets byIt is well defined: replacing $g$ by $gu$ for $u\in H\cap K$ changes neither component. It is injective becauseThe Cartesian product on the right has at most $nm$ elements. ThereforeThis proves the basic bound for a finite intersection of finite-index subgroups.

Put $L={\bigcap }_{i=1}^k{H}_i$. Repeated application of part (iv), or mathematical induction on $k$, givesPart (ii) therefore implies that $L$ is an infinite set. If $G\setminus U$ is nonempty, choose $g\notin U$. We claim its left coset $gL$ is wholly outside $U$. If $g\ell \in H_i$ for some $\ell \in L$, then $\ell \in H_i$ and closure under products and inverses would give $g=(g\ell ){\ell }^{-1}\in H_i$, a contradiction.

Thus $gL\subseteq G\setminus U$. Part (i) makes $gL$ infinite, so the complement is infinite. If there is no such $g$, the complement is empty. This proves the complement of a finite union of finite-index subgroups property:

Suppose there were only finitely many prime numbers, say $p_1,\dots ,p_k$. Work in the infinite group $(\mathbb{Z},+)$ and set $H_i=p_i\mathbb{Z}$. The left cosets are the $p_i$ residue classes modulo $p_i$, so each $H_i$ is a finite-index subgroup, with $|\mathbb{Z}/H_i|=p_i$.

By part (b)(v), $\mathbb{Z}\setminus {\bigcup }_i{p}_i\mathbb{Z}$ is empty or infinite. But part (a)(ii) identifies it as the nonempty two-element set $\{-1,1\}$. This contradiction proves Euclid's theorem:

Cosets and Lagrange. A left coset is $gH=\{gh:h\in H\}$. The left cosets of a subgroup partition a group, each with $|H|$ elements. Lagrange's theorem states $\boxed{|G|=[G:H]|H|}$ for a finite group, so every subgroup order divides the group order.

The 64-element group. Work in the commutative ring $R=\mathbb{Z}/4\mathbb{Z}$, which has zero divisors. Denote the displayed lower triangular matrix by $A(a,b,c)$, where $a,b,c$ are its three entries below the diagonal. Direct matrix multiplication givesAll arithmetic is modulo four, so the set is closed. Associativity comes from matrix multiplication, the identity element is $A(0,0,0)$, andThe formula works in either multiplication order and requires no division. Hence these matrices form a group of order $\boxed{|G|=4^3=64}$. This is the lower triangular version of the Heisenberg group over a commutative ring.

The abelian subgroup. If $a=c$, write $h(a,b)=A(a,b,a)$. ThenThe identity is $h(0,0)$ and the inverse is $h(-a,a^2-b)$, proving an abelian subgroup. Its two independent entries give $\boxed{|H|=16}$ and $\boxed{[G:H]=4}$, in accordance with Lagrange's theorem. This is the abelian plane subgroup of the Heisenberg group modulo four.

More explicitly, $A(a,b,c)\mapsto a-c$ is a surjective group homomorphism to the additive group of $R$, with kernel $H$. The four left cosets are exactly the four sets with fixed $a-c=t$, $t=0,1,2,3$; representatives are $A(t,0,0)$. Thus $H$ is also a normal subgroup.

Matrices squaring to the identity. The product formula givesThe first and third coordinates vanish only when $a,c\in \{0,2\}$. Then $ca=0$ modulo four, so the middle coordinate vanishes exactly when $b\in \{0,2\}$. Therefore the complete set isIt is a subgroup: multiplying two such elements keeps all three entries even, and their cross term $cd$ vanishes modulo four, so the product simply adds the three even coordinates. The identity is included and each element is its own inverse. In fact $K$ is isomorphic to the direct product of groups $C_2\times C_2\times C_2$. The seven nonidentity elements are involutions.

Finally, the last subset in the PDF is defined by $a\ne 0\ne b$, with no restriction on $c$. It has $3\cdot 3\cdot 4=36$ elements, which cannot be a subgroup because $36$ does not divide $64$. It also excludes the identity element. Thus

Definitions. A group homomorphism $\phi :G_1\to G_2$ preserves multiplication, $\phi (xy)=\phi (x)\phi (y)$. A group isomorphism is a bijective group homomorphism. The kernel of a group homomorphism isPreserving multiplication implies $\phi (e_1)=e_2$ and $\phi (x^{-1})=\phi (x{)}^{-1}$: the first follows by cancelling $\phi (e_1)$ from its square, and the second follows by applying $\phi$ to $x{x}^{-1}=e_1$.

Kernel subgroup. The identity belongs to $K$. For $x,y\in K$, $\phi (xy)=e_2{e}_2=e_2$, so $xy\in K$. Also $\phi (x^{-1})=e_2^{-1}=e_2$, so $x^{-1}\in K$. Thus $\boxed{K\le G_1}$.

Normality. For $g\in G_1$ and $k\in K$,Hence every $gk{g}^{-1}$ is again in $K$, proving $\boxed{K⊴G_1}$. Applying the same statement with $g^{-1}$ gives equality $gK{g}^{-1}=K$, and therefore $gK=Kg$.

Group of cosets. Define the product on the left cosets byTo check that this does not depend on representatives, replace $a,b$ by $a{k}_1,b{k}_2$. Thenbecause normality puts $b^{-1}{k}_{1}b$ in $K$. Thus the product is well defined. It is associative because multiplication in $G_1$ is associative; its identity is $K=e_{1}K$, and the inverse of $aK$ is $a^{-1}K$. Consequently the cosets form the quotient group $\boxed{G_1/K}$.

The map $aK\mapsto \phi (a)$ is well defined and gives an isomorphism $G_1/K\cong \mathrm{im}\phi$. If the original mapping is onto $G_2$, this becomes $G_1/K\cong G_2$. No surjectivity is needed for the kernel of a group homomorphism to be a normal subgroup or for the quotient group to exist.

Cosets and counting. A left coset of a subgroup $H$ in a group $G$ is $gH=\{gh:h\in H\}$. The left cosets partition $G$ and each has $|H|$ elements. Lagrange's theorem states that for a finite group, $\boxed{|G|=[G:H]|H|}$; in particular, the subgroup order divides the group order.

The matrix group. Work throughout in the finite field ${\mathbb{F}}_5$, so every nonzero entry has a multiplicative inverse. Denote a matrix by $A(a,b,c)$, with $a,c\ne 0$. Ordinary matrix multiplication, interpreted using modular arithmetic, givesIts diagonal entries remain nonzero, proving closure. Associativity is inherited from matrix multiplication, $A(1,0,1)$ is the identity element, andThis lies in the same set and multiplication checks the inverse in either order. Thus these matrices form a group. There are four choices each for $a$ and $c$ and five for $b$, giving $\boxed{|G|=4\cdot 5\cdot 4=80}$.

The subgroup. Write $h(a,c)=A(a,a-c,c)$. ThenIt contains the identity $h(1,1)$, is closed, and has inverse $h(a^{-1},c^{-1})$. Interchanging the two factors leaves the product unchanged because multiplication in ${\mathbb{F}}_5$ is commutative. Therefore $H$ is an abelian subgroup, withEquivalently $H$ is isomorphic to ${\mathbb{F}}_5^{\times }\times {\mathbb{F}}_5^{\times }$ by $h(a,c)\mapsto (a,c)$, which also explains its abelian property. Its order divides $80$, verifying Lagrange's theorem.

There are five distinct left cosets. Representatives can be taken as $U_t=A(1,t,1)$, $t\in {\mathbb{F}}_5$: indeedEvery matrix lies in exactly the coset specified by this unique $t$.

Elements whose square is the identity. Squaring givesThus $a,c\in \{1,-1\}=\{1,4\}$ and $b(a+c)=0$. For equal signs, $a+c\ne 0$ in ${\mathbb{F}}_5$, so $b=0$. For opposite signs, $a+c=0$ and all five $b$ values work. The complete set is thereforeIt contains $2+5+5=12$ elements, including the identity; the other eleven are involutions. Since $12$ does not divide $80$, Lagrange's theorem proves that this set is not a subgroup.