A sequence $(a_n)$ of real numbers is a convergent sequence with limit of a sequence $\ell$ whenDivergence to positive infinity meansIn both definitions the same $N$ must control every later term. The second condition is not convergence to a real number.

Enumerate both sets of peaks of a sequence in increasing order of index. If $a_n\to L$, the two selected subsequences both tend to $L$.

For the converse, suppose $a_{k_r}\to L$ and $a_{m_r}\to L$. Given $\epsilon >0$, choose $r$ large enough that both selected terms differ from $L$ by less than $\epsilon$. For every $n>\max \{k_r,m_r\}$, the two peak of a sequence conditions implyThe lower bound comes from the peak of a sequence of $(-a_n)$ at $m_r$. This is precisely the definition of $a_n\to L$. Thus convergence is equivalent to a common finite limit of a sequence for the upper and lower peak subsequences.

Choose $N$ beyond every peak of a sequence of both $(a_n)$ and $(-a_n)$. Every term with index at least $N$ has a strictly larger later term and a strictly smaller later term.

Suppose for a contradiction that $a_n\to L$. Starting at $a_N$, recursively select a strictly increasing subsequence and a strictly decreasing subsequence. Both must have limit of a sequence $L$. If their second terms are $u>a_N$ and $v<a_N$, respectively, passage to their limits of sequences givesa contradiction. Hence $(a_n)$ does not converge. This argument includes the possibility that either set of peaks of a sequence is empty.

Choose the sequenceAfter every index there are positive terms arbitrarily large and negative terms arbitrarily small. Thus every term has both a larger later term and a smaller later term. The former excludes a peak of a sequence $(b_n)$, and the latter excludes a peak of a sequence $(-b_n)$.

Choose the sequenceEvery even index is a peak of a sequence $(a_n)$, since its value is $1$ and all later values are at most $1$. Every odd index is a peak of a sequence $(-a_n)$ for the same reason. Both sets of indices are infinite.

If $a_n\to L$, every subsequence also has limit of a sequence $L$: its increasing indices eventually exceed any prescribed threshold. This applies in particular to every monotone subsequence.

Conversely, suppose $(a_n)$ is a bounded sequence and all its monotone subsequences converge to the same $L$. At least one exists by the monotone subsequence theorem. If $a_n$ did not tend to $L$, there would be ${\epsilon }_0>0$ and a subsequence $(a_{n_j})$ such thatApply the monotone subsequence theorem to $(a_{n_j})$. The resulting further subsequence is also a monotone subsequence of $(a_n)$, so it tends to $L$, contradicting the displayed bound. ThereforeThe bounded monotone sequence theorem ensures that these subsequences have finite limits of sequences; the hypothesis requires those values to agree.

It is useful to prove the contrapositive. Suppose $k$ is a peak of a sequence $(a_n)$ and $m$ is a peak of a sequence $(-a_n)$. The second condition says $a_j\ge a_m$ for every $j>m$. HenceOnly finitely many earlier terms remain, and their absolute values have a finite maximum. Together these give a bound for every term, so $(a_n)$ is a bounded sequence. Consequently,

The Bolzano-Weierstrass theorem states: every bounded sequence of real numbers has a convergent subsequence.

By the monotone subsequence theorem proved in the preceding part, a bounded sequence $(a_n)$ has a monotone subsequence $(b_r)$. It inherits the bounds of $(a_n)$. For completeness, the bounded monotone sequence theorem follows directly from completeness of the real numbers. If $(b_r)$ is increasing, let $L=\sup \{b_r:r\ge 1\}$. For every $\epsilon >0$, the definition of supremum supplies $r_0$ with $L-\epsilon <b_{r_0}\le L$. Thenso $b_r\to L$. If $(b_r)$ is decreasing, apply the increasing case to $(-b_r)$, or use its infimum. Thus in either case

Call $k$ a peak of a sequence when $a_k\ge a_m$ for every $m>k$. If there are infinitely many peaks of a sequence, list their indices in increasing order, $k_1<k_2<\cdots$. Since $k_{r+1}>k_r$,This is a decreasing subsequence; throughout this argument, a decreasing monotone sequence may have equal successive terms.

If there are only finitely many peaks of a sequence, choose $N$ beyond all their indices; take $N=1$ if there are none. Every $n\ge N$ fails to be a peak of a sequence, so some $m>n$ satisfies $a_m>a_n$. Starting with $n_1=N$, choose such an $n_{r+1}>n_r$ recursively. This givesThe two cases prove the monotone subsequence theorem without any boundedness hypothesis.

For a free group action, every stabilizer subgroup is $\{e\}$. By the orbit-stabilizer theorem, every orbit of a group action has cardinality $|G|$. Hence the two free group actions have the same numberof orbits of a group action. Choose one representative $s_i$ from each $\cdot$ orbit of a group action and one representative $t_i$ from each $\ast$ orbit of a group action, and match them using $1\le i\le r$.

Every $s\in S$ has a unique expression $s=g\cdot s_i$. The orbit of a group action determines $i$, and the free group action property determines $g$: if $g\cdot s_i=h\cdot s_i$, then $(h^{-1}g)\cdot s_i=s_i$, forcing $g=h$. Therefore the ruledefines a function $f:S\to S$ without ambiguity. For $h\in G$,which proves the required identity for every $s$.

Indeed, $f$ is a bijection: its inverse is $g\ast t_i\mapsto g\cdot s_i$, well defined because the second is also a free group action. If $S$ is empty, the unique empty function already satisfies the identity. Thus the result holds in every case, and proves the stronger statement that free finite actions with equal cardinality are isomorphic.

The orbits of a group action form a partition of a set $S$: every element lies in its own orbit of a group action, and two intersecting orbits of a group action coincide because each point can be moved back and then forward using inverse elements. By the orbit-stabilizer theorem, each orbit of a group action has cardinality dividing $8$, so its size is $1,2,4$ or $8$.

If every orbit of a group action had size $2,4$ or $8$, their sizes would add to an even number, contrary to $|S|=17$. Therefore there is a one-element orbit of a group action, say $\{s^{\prime }\}$. Every $g\in G$ sends $s^{\prime }$ into this same singleton, soThus $s^{\prime }$ is a fixed point of a group action. In fact the number of fixed points of a group action is odd, since all other orbits of a group action have even cardinality.

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.

A left group action of $G$ on a set $S$ is a function $G\times S\to S$, written $(g,s)\mapsto g\cdot s$, satisfyingfor all $g,h\in G$, $s\in S$. The orbit of a group action through $s$ and the stabilizer subgroup of $s$ are, respectively,The identity element belongs to $G_s$. If $g,h\in G_s$, then $(gh)\cdot s=g\cdot (h\cdot s)=g\cdot s=s$, so $gh\in G_s$. If $g\in G_s$, applying $g^{-1}$ to $g\cdot s=s$ gives $s=g^{-1}\cdot s$, so $g^{-1}\in G_s$. These facts prove that the stabilizer subgroup is a subgroup of $G$.

Choose distinct $a,b\ne e$, possible because $|G|>2$. By iii, $a$ and $b$ commute, and $a^2=b^2=e$. The four elementsare distinct: $ab=e$ would give $a=b$, while $ab=a$ or $ab=b$ would make $b=e$ or $a=e$. Commutativity and $a^2=b^2=e$ show that products stay in $H$, and every element is its own inverse. More explicitly,is a group isomorphism, with addition of exponents modulo two. Hence $G$ contains a subgroup isomorphic to the Klein four-group.

The converse is false. The direct product of groups $C_2\times C_2\times C_3$ contains $C_2\times C_2\times \{e\}$, but it has elements of order of a group element three. Thus the existence of the specified subgroup does not force every nonidentity element to have order of a group element two.

The hypothesis implies $x^{-1}=x$ for every $x\in G$, including $e$. For $a,b\in G$,Thus $G$ is an abelian group. This is the group of exponent two is abelian argument.

The converse is false: the cyclic group $C_3$ is an abelian group, but its two nonidentity elements have order of a group element three.

If $G=⟨g⟩$, any two elements are $g^a,g^b$, andThus every cyclic group is an abelian group.

For the converse, the Klein four-group $C_2\times C_2$ is an abelian group, with componentwise multiplication. Each nonidentity element has order of a group element two, so none generates its four elements. It is therefore not a cyclic group.

Let $|G|=p$, with $p$ a prime number, and choose $g\ne e$. By Lagrange's theorem, the order of a group element $g$ divides $p$ and is greater than one. It is therefore $p$, and $G=⟨g⟩$. Hence a group of prime cardinality is a cyclic group.

The converse is false: the cyclic group $C_4$ has four elements, and $4$ is not a prime number.

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.

For $a\in A$, the element $a=ae$ belongs to both $A$ and $aA$. The assumed equal-or-disjoint property therefore gives $aA=A$.

If $a,b\in A$, then $ab\in aA=A$, so $A$ is closed under multiplication. Also $e\in A=aA$, so some $c\in A$ satisfies $ac=e$. Therefore $c=a^{-1}\in A$. Along with the given identity element, closure and existence of inverse elements establish that $A$ is a subgroup of $G$. This is the subgroup test. The proof even works without assuming that $G$ is finite.

The order of a group element $g$ is the smallest positive integer $d$ with $g^d=e$. Such an integer exists in a finite group: two powers in $e,g,g^2,\dots$ must agree, and cancellation gives a positive power equal to $e$.

By minimality, $e,g,\dots ,g^{d-1}$ are distinct. Every integer power of $g$ is one of them, by division of its exponent by $d$. They form the cyclic subgroupApplying Lagrange's theorem to this subgroup gives