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A group is a nonempty set with a closed binary operation that is associative, has an identity element, and gives every element an inverse element. A subgroup is a subset that is itself a group under the inherited operation. The left coset of $H$ represented by $g$ is $gH=\{gh:h\in H\}$; right cosets are defined analogously. Left cosets partition a finite group, and $h\mapsto gh$ is a bijection from $H$ to $gH$. Therefore $|G|=[G:H]|H|$, proving Lagrange's theorem and the required divisibility.

For the given matrices, put $a=A$ and $c=AB=iI$. Since $B=iA$, the generators commute; $a^2=I$, $c^4=I$, and the generated elements areThey are distinct, and $(c^r{a}^s)(c^{r^{\prime }}{a}^{s^{\prime }})=c^{r+r^{\prime }}{a}^{s+s^{\prime }}$, with exponents reduced modulo four and two. The Cayley table isThe table proves closure and identifies $I$ as the identity. The inverse element of $c^r{a}^s$ is $c^{-r}{a}^s$, and associativity follows from matrix multiplication. Thus this is the direct product of cyclic groups of orders four and two, an abelian group of order eight.

Its subgroups are exactlyTo see the list is complete, Lagrange's theorem leaves orders one, two, four and eight. There are exactly three elements of order two, giving the three order-two subgroups. An order-four subgroup containing an element of order four must be one of the two cyclic groups displayed; without such an element it must be the four-element set of elements whose squares are $I$. This is the enumeration in subgroups of C4 times C2.

For example choose $H=\{I,c,c^2,c^3\}$. A complete set of cosets is $\boxed{H,aH=\{a,ca,c^{2}a,c^{3}a\}}$. Since $G$ is Abelian, the left and right cosets coincide.

Observe that $A=-I$, $B^2=I$ and $C=-B$. Thus $A^2=B^2=C^2=I$, while $AB=BA=C$, $AC=CA=B$ and $BC=CB=A$. The multiplication table, with the left factor labeling the row, isIt proves closure of the set. Associativity is inherited from matrix multiplication, $I$ is the identity element, and every element is its own inverse element. Therefore this set is a group. The table is symmetric, so it is an Abelian group; its three nonidentity elements have order two, identifying it as the Klein four-group $C_2\times C_2$.

Let $T$ be the set of torsion elements of the given Abelian group. It contains the identity element, which has order one. If $g,h\in T$ have orders $m,n$, set $\ell =\mathrm{lcm}(m,n)$. Commutativity permits powers of the product to split:Hence $g{h}^{-1}$ has finite order, dividing $\ell$, and lies in $T$. The nonempty set $T$ therefore satisfies the subgroup test and is a subgroup:This is the torsion subgroup. The proof uses commutativity exactly when rearranging the factors in $(g{h}^{-1}{)}^{\ell }$; one cannot use that step in an arbitrary noncommutative group. The subgroup need not itself be a finite set.

The hypothesis implies $g^2=e$ for every $g\in G$, including the identity, so $g^{-1}=g$. For arbitrary $a,b\in G$, their product also equals its own inverse. The inverse-of-a-product rule therefore givesThus $G$ is an abelian group. Finiteness is not needed for this implication; it holds for every group in which each element squares to the identity.

By Lagrange's theorem, the possible values of the order of a group element areThese are possibilities across groups of order eight; they need not all occur in one group.

If every element has order at most two, every element equals its own inverse. Thus for any $x,y$,so $G$ is an abelian group. Choose $a\ne e$, and then $b\notin ⟨a⟩$. The four elements $e,a,b,ab$ are distinct and form a subgroup $H$: commutativity and $a^2=b^2=e$ prove closure and inverses. Choose $c\notin H$. The elements of $H$ and $Hc$ are disjoint, since $hc\in H$ would imply $c\in H$. These two four-element sets therefore exhaust $G$.

Every element consequently has a unique expression $a^i{b}^j{c}^k$, with $i,j,k\in \{0,1\}$. The map from triples to these products is bijective and respects multiplication by addition of exponents modulo two. Hencea direct product of groups and an elementary abelian group.

The sum of two odd integers plus one is odd, so the operation is closed. Both associations of a triple give $a+b+c+2$, proving associativity. Its identity element is $-1$, and the inverse element of $a$ is $-a-2$, also odd, since $a+(-a-2)+1=-1$. Thus it is an abelian group.

More explicitly, shifted addition on odd integers is ordinary integer addition in different coordinates:This bijection is a group isomorphism to $(\mathbb{Z},+)$.

The requested two-dimensional irreducible representation does not exist for the Vierergruppe. The Klein four-group $V_4$ is an abelian group in which every nonidentity element has order two. Its complex irreducible representations are all one-dimensional. The same holds over the real numbers here: the commuting representing involutions are simultaneously diagonalizable with real eigenvalues $\pm 1$, so every two-dimensional representation has an invariant line. Each genuine irreducible character satisfies $\chi (g^2)=\chi (1)=1$, and hence has Frobenius-Schur indicator $\nu =1$.

The printed notation $C_4$ suggests a different possible interpretation, which must not be conflated with the Klein four-group. For the cyclic group of order four, a real two-dimensional irreducible representation is generated byIts character on $1,J,-I,-J$ is $(2,0,-2,0)$, giving $\nu =(2-2+2-2)/4=0$. Over the complex numbers this representation splits into two conjugate one-dimensional characters, so it is not a two-dimensional complex irreducible representation. These alternatives explain what can be computed without silently changing the printed group or the ground field.

The symmetric group $S_3$ is generated by transpositions, each of order two. A group homomorphism to ${\mathbb{Z}}_3$ must map every transposition to zero, because the only target element annihilated by two is zero. Hence the map to ${\mathbb{Z}}_3$ is necessarily trivial.

For $S_3\to {\mathbb{Z}}_2$, the 3-cycles map to zero because their order divides three. All transpositions are conjugate group elements, and a map to an abelian group takes conjugate elements to the same value. That common value is either zero or one. Both choices work: they give the trivial map and the sign homomorphism, with an odd permutation sent to $1$ and an even permutation sent to $0$.

Use the vector-space axioms: addition is an abelian group operation with identity zero; ${\lambda }_1(x)=x$; ${\lambda }_a{\lambda }_b(x)={\lambda }_{ab}(x)$; ${\lambda }_a(x+y)={\lambda }_a(x)+{\lambda }_a(y)$; and ${\lambda }_{a+b}(x)={\lambda }_a(x)+{\lambda }_b(x)$, with ${\lambda }_0(x)=0$. Group inverses may be expressed by ${\lambda }_{-1}$, so no extra inverse symbol is needed.

Add the dimension-unbounded schemeOver a finite field, a finite-dimensional space has finitely many Elements, while every infinite-dimensional space is infinite. Hence this scheme axiomatizes exactly the infinite underlying vector spaces requested. In this finite-field case the vector-space axiom list itself is finite; the infinitude of the additional dimension scheme is substantive.

Yes. As an Abelian group, the reals are a nontrivial torsion-free divisible group, equivalently a vector space over the rationals. Rational scalar multiplication is uniquely definable by equations such as $ny=mx$, so adding those operations is a definitional expansion.

In this language, quantifier elimination for torsion-free divisible Abelian groups reduces formulas to Boolean combinations of rational linear equations. Over finitely many parameters, a complete one-variable type either fixes a value in their linear span or excludes every value in that span. The first kind is realized in the span itself; the second is realized by any real number outside it. That span is countable, while the reals are uncountable. Thus every such type is realized and the additive real group is an omega-saturated model.

For finite tuples, the same argument successively supplies any required finite number of new rationally linearly independent elements. The order and multiplication of the real field are absent from this language, so they cannot impose additional type conditions.

An integral basis is a $\mathbb{Z}$-basis of the ring of integers ${\mathcal{O}}_K$; its elements also form a $\mathbb{Q}$-basis of $K$.

To prove existence, first multiply the elements of a rational basis by suitable nonzero integers to obtain an integral rational basis $w_1,\dots ,w_n$. Put $M={\sum }_j\mathbb{Z}{w}_j$. The trace pairing is nondegenerate in characteristic zero, and the matrix $T=(\mathrm{Tr}(w_i{w}_j))$ has integer entries and nonzero determinant $\Delta$. If $\alpha ={\sum }_j{c}_j{w}_j$ is integral, all $\mathrm{Tr}(\alpha w_i)$ are integers, so $Tc\in {\mathbb{Z}}^n$. The adjugate formula gives $c\in {\Delta }^{-1}{\mathbb{Z}}^n$. ConsequentlyA subgroup of a free abelian group is free, and these inclusions force rank $n$. A $\mathbb{Z}$-basis of ${\mathcal{O}}_K$ is therefore an integral basis.

Now put $s={\theta }^2=\sqrt{d}$ and $F=\mathbb{Q}(s)$. The polynomial $X^4-d$ is irreducible by Eisenstein at any prime dividing the squarefree $d$, except for $d=-1$, where $(X+1{)}^4+1$ is Eisenstein at $2$. Thus $[K:\mathbb{Q}]=4$ and $[K:F]=2$. Every $\gamma \in K$ has the unique form $\alpha +\beta \theta$ with $\alpha ,\beta \in F$. The nontrivial $F$-automorphism sends $\theta$ to $-\theta$, so an integral $\gamma$ has integral conjugate $\alpha -\beta \theta$. Relative trace and norm of integral elements are integral in $F$. Applying them to $\gamma$ and taking the trace of $\theta \gamma$ givesThe power basis has discriminantIndeed, the resultant of $X^4-d$ and $4X^3$ is $4^4\prod {\theta }_j^3=-256d^3$; the discriminant sign factor is $(-1{)}^6=1$. By the discriminant-index formula for an integral lattice, a prime dividing $[{\mathcal{O}}_K:\mathbb{Z}[\theta ]]$ must divide $2d$.

No odd prime $p\mid d$ can divide this index. The Eisenstein equation gives ${\theta }^4/p\in {\mathcal{O}}_K$, while $r{\theta }^i/p$ has nonintegral field norm for $p\nmid r$ and $0\le i<4$: its $p$-adic norm valuation is $i-4<0$. If $p^{-1}{\sum }_{i=0}^3{r}_i{\theta }^i$ were integral, multiplication by ${\theta }^3$ first forces $p\mid r_0$; subtracting $r_0/p$ and multiplying by ${\theta }^2$ forces $p\mid r_1$. Continuing forces all coefficients divisible by $p$. This excludes an index element of order $p$.

To exclude $2$, any index element of order two would be represented by an integralwith integers $a,b,c,e$ not all even. Here $\alpha =(a+cs)/2$ and $\beta =(b+es)/2$. The integrality of ${\alpha }^2-{\beta }^{2}s$ forcesSince $d\equiv 3(\mathrm{mod}4)$, the first congruence implies $a,c$ have the same parity. If both were odd, the second would require $b^2+3e^2\equiv 2(\mathrm{mod}4)$, impossible. Hence $a,c$ are even. The second congruence then forces $b,e$ to have the same parity, while the first forces $be$ even. Thus $b,e$ are even too, a contradiction. Therefore the index is one andFor the last request, let $t=\sqrt[4]{5}$ and take $\delta =(t+t^3)/2$. It satisfies ${\delta }^2=(5+3\sqrt{5})/2$, and henceThis monic integer polynomial proves integrality. Its coordinates in the rational power basis have halves, so $\delta \notin \mathbb{Z}[t]$. Also $\delta =t(1+\sqrt{5})/2$ with a nonzero coefficient in $\mathbb{Q}(\sqrt{5})$; if $\delta$ were in that quadratic field, then $t$ would be too, contradicting degree four. Thus it meets both exclusions.

An integral basis is a $\mathbb{Z}$-basis of the ring of integers ${\mathcal{O}}_K$; it consists of $n$ algebraic integers and is also a $\mathbb{Q}$-basis of $K$.

The lattice $M$ has finite index $h=[{\mathcal{O}}_K:M]>1$. Choose a prime $p\mid h$. The finite abelian group ${\mathcal{O}}_K/M$ has an element of order $p$, represented by $\alpha \in {\mathcal{O}}_K\setminus M$. ThusIf all $c_j$ were divisible by $p$, cancellation in the characteristic-zero field would put $\alpha$ in $M$, a contradiction. The discriminant-index formula for an integral lattice giveswhere ${\Delta }^2(K)$ is an integer. Hence $p^2\mid {\Delta }^2(M)$, and the required integral element is $\alpha =p^{-1}{\sum }_j{c}_j{w}_j$.

A solvable group has a finite subnormal series from the identity to the group with abelian group factors. An equivalent definition uses the derived series: set $G^{(0)}=G$ and $G^{(j+1)}=[G^{(j)},G^{(j)}]$; the series must reach the identity. For completeness, an abelian-factor series forces successive derived subgroups into its successive terms, while a terminating derived series itself supplies such a series.

If $H\le G$, induction gives $H^{(j)}\le G^{(j)}$: every commutator of elements of $H^{(j)}$ is also a commutator in $G^{(j)}$. Thus termination for $G$ implies termination for $H$, proving that every subgroup of a solvable group is solvable.

In $S_4$, useThe Klein four-group is invariant under conjugation in $S_4$; the other normality assertions follow from the alternating group being the kernel of the sign map. The quotient group factors are $V_4,C_3,C_2$, all abelian. Hence $S_4$ is solvable.

For the last example take $A_5\times C_2$. Its subgroup $\{1\}\times C_2$ is nontrivial, proper and normal, so the product is not a simple group. Projection onto $A_5$ is surjective, and every quotient of a solvable group is solvable, because a homomorphism carries each derived subgroup onto the derived subgroup of its image. But the nonabelian simple group $A_5$ has $A_5^{\prime }=A_5$: its derived subgroup is nontrivial and normal. Thus the product is not solvable either. One standard justification of the simplicity used here is that the nonidentity conjugacy class sizes of $A_5$ are $15,20,12,12$; no proper nontrivial union with the identity has size dividing $60$, as a normal subgroup would require by Lagrange's theorem for finite groups.

A composition series is a finite chainwhose inclusions are strict and whose quotients $G_i/G_{i-1}$ are simple. These quotients are the composition factors. The normal subgroup condition is relative to the next member, rather than necessarily to all of $G$.

Induct on the order of a finite group. The trivial group has the empty composition series. Otherwise choose a maximal normal subgroup $N⪇G$, which exists because there are finitely many subgroups. The subgroup correspondence theorem identifies normal subgroups of $G/N$ with normal subgroups of $G$ containing $N$, so maximality makes $G/N$ a simple group. By induction $N$ has a composition series; appending $G$ gives one for $G$.

For example, the nonabelian symmetric group $S_3$ haswith composition factors $C_3$ and $C_2$, both abelian.

Write $C_n=⟨g⟩$ and set $N=⟨g^p⟩$. The order of $g^p$ is $n/\gcd (n,p)=m$, so $N\cong C_m$. Every subgroup of an abelian group is a normal subgroup. The quotient group $C_n/N$ is generated by $gN$ and has order $n/m=p$, hence is $C_p$ and is a simple group. By the subgroup correspondence theorem, there is no proper normal subgroup strictly between $N$ and $C_n$. Therefore $N$ is a maximal normal subgroup with the required quotient group.

For nonempty finite subsets $A,B$ of an abelian group, $|A+B|\le K|A|$ implies that some nonempty $X\subseteq A$ satisfies this bound simultaneously for every integer $m\ge 0$, with $0B=\{0\}$. The Petridis minimal-growth lemma proves this by choosing $X$ to minimize $|X+B|/|X|$. The Ruzsa triangle inequality then gives the Plünnecke-Ruzsa inequality $|kB-\ell B|\le K^{k+\ell }|A|$.

For a finite nonempty subset of an abelian group with doubling constant at most $K$, the Plünnecke-Ruzsa inequality gives $|3A-2A|\le K^5|A|$. Applying the Ruzsa covering lemma to $2(A-A)$ using $A$ gives $2T\subseteq D+T$ for $T=A-A$ and $|D|=q\le K^5$. Thus $\ell T\subseteq (\ell -1)D+T$, and counting multiplicities in the finite set $D$ proves the displayed polynomial bound. For fixed $K>1$, it is eventually at most $K^{ϵ\ell }|A|$ for every $ϵ>0$.

A category is preadditive if every hom-set is an abelian group and composition is additive in each variable. Zero morphisms and differences of parallel morphisms are therefore defined. A zero object or finite biproducts need not exist; these are extra requirements in an additive category.

If $S^n\to N$ has prime degree $p$ and $N$ is a closed connected oriented $n$-manifold, every integral homology group of $N$ in degrees strictly between zero and $n$ is a finite $p$-primary abelian group. Consequently one power of $p$ annihilates all these groups. The proof combines Poincare duality over each field ${\mathbb{F}}_q$, $q\ne p$, with the universal coefficient theorem for homology and finite generation.

If $K/{\mathbb{Q}}_p$ is finite of degree $d$, then as a topological abelian groupwhere ${\mu }_{p^a}(K)$ is the finite group of all roots of unity in $K$ whose orders are prime powers dividing a power of $p$.