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For an $n$-dimensional random vector with a density and finite differential entropy in bits, define its entropy power by $N(X)=(2\pi e{)}^{-1}{2}^{2h(X)/n}$. The entropy power inequality states that for independent such random vectors,Here the sum's differential entropy must be defined, with $+\infty$ permitted.

Since $g^{\prime }(x)>0$, the mean value theorem makes $g$ strictly increasing. Its image is an interval $I$, and its inverse on $I$ has derivative $1/g^{\prime }(g^{-1}(y))$. The density change of variables givesThis does not require $g$ to map onto all of $\mathbb{R}$. Taking logarithms at $y=g(X)$ and then expected values proves the differential entropy under an increasing transformation:The two finite expectations in the assumptions justify splitting the expectation and imply that the transformed differential entropy is finite.

To derive the product inequality, assume that $h(Y_i)$ and ${\mu }_i=\mathbb{E}{\log }_2{Y}_i$ are finite and that the differential entropy of the logarithmic sum is defined. Set $Z_i=\ln Y_i$, using natural logarithms for this transformation while measuring differential entropy in bits. On the positive half-line the derivative of $\ln y$ is $1/y$, so the same density argument givesThe $Z_i$ are independent random variables, and the one-dimensional entropy power inequality yieldsSince $Y_1{Y}_2=\exp (Z_1+Z_2)$, another change of variables givesMultiplying the entropy power inequality by $2^{2({\mu }_1+{\mu }_2)}$ proves the multiplicative entropy power inequality:

The printed positivity and density assumptions alone do not ensure that the quantities in this last formula exist. For a concrete example, let $T$ have the standard Cauchy distribution and put $Y=e^T$. Thenis a density, but $\mathbb{E}{\log }_{2}Y=({\log }_{2}e)\mathbb{E}T$ is undefined because its positive and negative parts both diverge. Two independent copies satisfy the printed hypotheses but leave ${\alpha }_1,{\alpha }_2$ undefined. The product inequality therefore needs the finiteness or well-definedness conditions used above.

The requested representation is the left coset actionIt is well defined because equal cosets remain equal after multiplication by $g$, and $\rho (g{g}^{\prime })=\rho (g)\rho (g^{\prime })$. It is transitive. For $H=1$ it is the left regular action; for general $H$ it is the associated transitive coset representation.

Its kernel is the normal core of a subgroupIndeed, $g$ fixes every $xH$ precisely when $x^{-1}gx\in H$ for every $x$. Thus $N⊲G$ and $N\le H$. If $[G:H]=i<\infty$, the image lies in the finite symmetric group $S_i$, soIn particular $N$ has finite index in $H$.

For Higman group as an amalgam of iterated HNN extensions, use the cyclic squaring presentationFirst form $L=⟨b,c\mid b^{-1}cb=c^2⟩$, an HNN extension of the infinite cyclic group $⟨c⟩$. Its base embeds, so $c$ has infinite order; $b$ has infinite order as well, since the homomorphism $L\to \mathbb{Z}$ sending $b\mapsto 1,c\mapsto 0$ is onto. Consequently $⟨b⟩$ and $⟨b^2⟩$ are isomorphic infinite cyclic subgroups. Adjoining the stable letter $a$ with $a^{-1}ba=b^2$ givesWe claim $⟨a,c⟩$ is a rank-two free group. No nonzero power of $c$ belongs to $⟨b⟩$: applying the map $b\mapsto 1,c\mapsto 0$ would make that power equal to $1$, contradicting the infinite order of $c$. Thus no nonzero power of $c$ belongs to $⟨b^2⟩$ either. Any nonempty freely reduced word in $a,c$ that contains $a$ is therefore a reduced HNN sequence: between inverse $a$ letters, a nonzero $c$ power cannot create a pinch. Britton's lemma makes it nonidentity. A word containing only $c$ is nonidentity by the base embedding. This proves the claim.

Likewiseis obtained from $⟨d,a\mid d^{-1}ad=a^2⟩$ by adjoining $c$, and its subgroup $⟨c,a⟩$ is free of rank two by the same argument. Form the amalgamated free product $P{\ast }_{⟨a,c⟩}Q$, identifying these two free subgroups generator by generator. Its presentation is exactly that of $\mathcal{H}$. The component embeddings allowed in the question show that it contains $P$, hence is infinite. It has four group generators and four relators, so it is finitely presented.

To show that this Higman group has no nontrivial finite quotient, consider any finite image and the orders of the four generator images. A relation $x^{-1}yx=y^2$ makes $y$ and $y^2$ have the same order, so every generator order is odd. Suppose one is nontrivial, and let $p$ be the smallest prime dividing any of the four orders. Choose a generator $y$ whose order is divisible by $p$, with predecessor $x$ satisfying the displayed relation. Iterating conjugation $|x|$ times givesThe multiplicative order $d$ of $2$ modulo $p$ thus divides $|x|$. Also $d\mid p-1$ and $d>1$, since $2\not\equiv 1(\mathrm{mod}p)$. Any prime divisor of $d$ is smaller than $p$ and divides the order of $x$, contradicting minimality of $p$. Hence every generator image is trivial, so every finite image is trivial. A proper finite-index subgroup would give a nontrivial transitive finite coset image, which is impossible. Therefore $\mathcal{H}$ is infinite and finitely presented, with no proper finite-index subgroups.

A maximal normal subgroup is a proper normal subgroup $M⊲G$ with no normal subgroup strictly between $M$ and $G$. To prove that finitely generated groups have maximal proper normal subgroups, fix a proper normal subgroup $K$ and order the proper normal subgroups containing $K$ by inclusion. For a chain, its union is normal and contains $K$. It is still proper: if the union were all of $G$, each element of a finite generating set would lie in some chain member, and the chain order would put all these finitely many group generators in one member. That member would equal $G$, a contradiction. The Zorn lemma therefore supplies a maximal proper normal subgroup $M$ containing $K$.

Apply this to $\mathcal{H}$ and $K=1$. By the normal-subgroup correspondence, $\mathcal{H}/M$ is nontrivial and simple. It is finitely generated as a quotient of $\mathcal{H}$. It cannot be finite, since $\mathcal{H}$ has no nontrivial finite quotients. Thus

The final assertion for arbitrary proper subgroups is false. Take $G=S_3$ and $H=⟨(12)⟩$. If a normal subgroup contained $H$, it would contain all conjugate transpositions, which generate $S_3$, so it would equal $G$. Hence this proper subgroup lies in no proper normal subgroup, and in particular in no maximal normal subgroup. The maximal-normal existence result requires the starting subgroup itself to be normal.

A group $G$ is a finitely presented group if there is an epimorphism $F(X)\to G$ from a finite-rank free group whose kernel is the normal closure of finitely many words $R$. Equivalently, $G=⟨X\mid R⟩$ with both $X$ and $R$ finite. Normal closure means that every kernel element is a finite product of conjugates of the chosen relators or their inverses.

For independence of finite presentation from finite generators, suppose $G=⟨y_1,\dots ,y_m\mid r_1,\dots ,r_s⟩$, and let $x_1,\dots ,x_k$ be any other finite generating set. Choose words $u_j(X)$ representing $y_j$ and words $v_i(Y)$ representing $x_i$. Consider the finite presentationSending each $x_i$ to its actual element gives a surjection $P\to G$. Conversely, send $y_j$ to $u_j(X)$; the substituted relators make this a homomorphism $G\to P$. Their composition on $G$ fixes every $y_j$, and their composition on $P$ fixes every $x_i$ by the extra relations. They are inverse isomorphisms. Hence finite presentability is independent of the chosen finite generating set.

For subgroups $A,B\le G$ and an isomorphism $\varphi :A\to B$, adopt the conventiona quotient of the free product $G\ast ⟨t⟩$. The generator $t$ is the stable letter of this HNN extension. Fix right coset transversals $R_{+}$ for $B\backslash G$ and $R_{-}$ for $A\backslash G$, each with $1$ representing the subgroup itself. Thus each $g$ has a unique decomposition $g=br$ with $b\in B,r\in R_{+}$, and a unique decomposition $g=ar$ with $a\in A,r\in R_{-}$.

A normal form is a stringwhere $r_i\ne 1$ whenever ${ϵ}_{i+1}=-{ϵ}_i$. The case $n=0$ is just $g_0$. The prohibition removes adjacent inverse stable letters that would cancel. The following permutation proof of HNN normal form uniqueness proves both existence and uniqueness.

Let $\mathcal{N}$ be the set of these formal strings. A base element $g$ acts by left multiplication on the initial coefficient $g_0$; call this permutation $L_g$. To define $L_t$, decompose the initial coefficient as $g_0=br$, with $b\in B,r\in R_{+}$. Unless $r=1$ and the old first stable letter is $t^{-1}$, replace the initial segment byIn the exceptional case, if the old string begins $b{t}^{-1}{r}_1\cdots$, cancel the new $t$ with that $t^{-1}$ and obtain the string beginning ${\varphi }^{-1}(b)r_1\cdots$. These operations use the defining identity $tb={\varphi }^{-1}(b)t$.

Define $L_{t^{-1}}$ similarly: write $g_0=ar$ with $a\in A,r\in R_{-}$, prepend $\varphi (a)t^{-1}r$, and cancel with an old first $t$ exactly when $r=1$. The resulting strings satisfy the normal-form conditions. These two maps are inverses. If the first operation inserts a letter, the reverse operation decomposes its new initial coefficient in the associated subgroup and cancels that letter. If the first operation cancels a letter, the reverse coset decomposition recovers the removed representative and reinserts it; an additional unwanted cancellation is excluded by the original normal-form condition.

The permutations satisfy $L_g{L}_h=L_{gh}$ andIndeed, in the decomposition $g_0=br$, multiplying $g_0$ on the left by $\varphi (a)$ leaves its representative $r$ unchanged and changes $b$ to $\varphi (a)b$. Its image under ${\varphi }^{-1}$ is $a{\varphi }^{-1}(b)$, exactly the coefficient produced by multiplying the normalized result on the left by $a$. The cancellation case has the same representative and the same conclusion. ThereforeAll defining relations of the HNN extension hold in this action, so it induces a homomorphism from $G{\ast }_{\varphi }$ into the permutations of $\mathcal{N}$.

Starting with the empty string, successive left multiplications normalize any word, and every normalization step uses a defining relation. This proves existence. If two normal words represented the same group element, their induced permutations would agree. Applied to the empty string, each normal word gives exactly its own formal string, since its representatives are already chosen and all forbidden cancellations have been excluded. Thus the strings are equal. This proves the normal form theorem for an HNN extension, including the injectivity of the base group $G\to G{\ast }_{\varphi }$.

A reduced sequence in an HNN extension is a wordwith no pinch $t^{-1}at$ for $a\in A$ and no pinch $tb{t}^{-1}$ for $b\in B$. Equivalently, if adjacent exponents are $-1,+1$, their intervening coefficient is outside $A$; if they are $+1,-1$, it is outside $B$.

Normalize such a word from right to left using the procedure above. Inductively its stable-letter length does not change. To see why, suppose the next letter has exponent $\delta$. Moving an associated-subgroup factor through it contributes to its left an element of ${\varphi }^{-1}(B)=A$ when $\delta =1$, or of $\varphi (A)=B$ when $\delta =-1$. If the preceding exponent is $-\delta$, this factor belongs to that preceding letter's pinch subgroup. Multiplying an intervening coefficient outside that subgroup by an element inside it cannot put the product inside the subgroup. Hence no cancellation occurs; when the two exponents agree, cancellation is impossible anyway. Thus a reduced word with $n\ge 1$ has a normal form still containing $n$ stable letters, which cannot be the identity normal form. This is Britton's lemma:

For the final example, write the centralizing extension asIt is generated by $a,b,t$. Suppose it were finitely presented. By the finite-generator independence already proved, the kernel of $F(a,b,t)\to E$ would be normally generated by finitely many words. Each of these kernel group generators is a finite product of conjugates of the displayed defining relators, so only finitely many indices occur in all their expressions. Choose $N$ at least as large as all of them. The full kernel would then equal the normal closure of the first $N$ relators: one inclusion follows from those expressions, and the reverse inclusion follows because all these relators already belong to the kernel.

But the group with only those first $N$ relators is the identity HNN extension over $H_N=⟨w_1,\dots ,w_N⟩$. Since the $w_j$ are a free basis of $H$, $w_{N+1}\notin H_N$. The word $t^{-1}{w}_{N+1}t{w}_{N+1}^{-1}$ is reduced in this partial extension, so Britton's lemma makes it nonidentity there. It is identity in $E$, contradicting equality of the kernels. ThereforeThis is the free-basis instance of the general fact that finite presentation of a centralizing HNN extension detects finite generation of its associated subgroup.

Choose free group generators $a_1,\dots ,a_n$. Construct the Schreier coset graph $\Gamma$ with vertices the right cosets $Hg$ and a positively oriented edge labelled $a_k$ from $Hg$ to $Hg{a}_k$. This is well defined and connected. Its inverse edge carries $a_k^{-1}$. A word in the group generators determines a unique path from the base vertex $H$; that path closes exactly when the word represents an element of $H$.

Deleting a consecutive edge followed by its inverse corresponds exactly to free cancellation in the word. Since reduced words are unique in a free group, the resulting map from based loops modulo these cancellations to $H$ is injective as well as surjective, and respects concatenation. Thus $H$ is the fundamental group of $\Gamma$. More explicitly, choose a spanning tree $T$ and, for each unoriented edge outside $T$, choose one orientation. The tree path from the base vertex to that edge, followed by the edge and the returning tree path, defines an element of $H$. Every loop is a product of these elements and their inverses: insert the appropriate tree paths between successive edges, and tree edges contribute trivial factors. Collapsing the tree leaves one loop for each remaining edge. A product of the chosen basis loops maps to the corresponding word in these loop labels. If the original path could cancel to the constant path, its image could do so as well, since collapsing tree edges preserves every backtracking cancellation. A nonempty reduced word in the remaining labels cannot cancel to the empty word, proving that there are no further relations. These elements therefore form a free basis. This proves that $H$ is a free group without assuming the subgroup theorem.

If $i=[F_n:H]$ is finite, the graph has $i$ vertices and $ni$ unoriented edges, counted by their positive labels; loops and parallel edges count separately. A spanning tree has $i-1$ edges. The Nielsen–Schreier formula follows:For a sequence of finite indices $i_j\to \infty$, this gives the rank-to-index limit for subgroups of a free group

Finally, let $\theta :F_n\to F_n$ be an endomorphism, and put $L=\theta (F_n)$. It is generated by the $n$ images $\theta (a_k)$. If its index $d$ is finite, it is free of rank $r=1+d(n-1)$. A rank-$r$ free group needs at least $r$ group generators: its abelianization is ${\mathbb{Z}}^r$, and reduction modulo $2$ gives the $r$-dimensional vector space $(\mathbb{Z}/2\mathbb{Z}{)}^r$, which cannot be generated by fewer than $r$ vectors. ThereforeSince $n\ge 2$, this forces $d\le 1$, hence $L=F_n$. This proves that a finite-index image of a free-group endomorphism is surjective:

To compute the rank of an elliptic curve, translate the rational point of order two to $(0,0)$ and put the curve in the form $y^2=x^3+a{x}^2+bx$, with $a,b\in \mathbb{Z}$ after clearing denominators and $b(a^2-4b)\ne 0$. Its two-isogenous partner and the elliptic isogeny areThe maps extend over their exceptional points. The dual isogeny isThe two-torsion square-class homomorphism is $\alpha :E(\mathbb{Q})\to {\mathbb{Q}}^{\times }/{\mathbb{Q}}^{\times 2}$, defined by $\alpha (O)=1$, $\alpha ((0,0))=[b]$, and $\alpha ((x,y))=[x]$ otherwise. Define ${\alpha }^{\prime }$ similarly on $E^{\prime }$ with $b^{\prime }=a^2-4b$.

Here are the ingredients behind the descent procedure. For a generic chord $y=mx+c$, the cubic of intersection coordinates has root product $c^2$. As the third intersection is the negative of the group sum and negation leaves $x$ unchanged, this proves the square-class homomorphism property. For a line through $(0,0)$, the two nonzero intersection coordinates have product $b$, so the assigned class $[b]$ gives the same multiplicativity. A vertical line has two equal coordinates and contributes their square, and $O$ has identity class. These observations also cover tangencies and the exceptional sums. Its kernel is $\hat{\varphi }(E^{\prime }(\mathbb{Q}))$. Indeed, the displayed formula gives $x(\hat{\varphi }(X,Y))=(Y/(2X){)}^2$. Conversely, if a point on $E$ has $x=r^2\ne 0$, the possible preimage coordinates satisfyIts elliptic-curve discriminant is $16(r^4+a{r}^2+b)=16y^2/r^2$, a rational square. A root $X$ and $Y=2rX$, with the sign chosen appropriately, give a preimage. The identity is always in the image. The point $(0,0)$ is in the image exactly when $b$ is a square: its nonzero preimage coordinates are $X=a\pm 2\sqrt{b}$, with $Y=0$. Thus the exceptional points satisfy precisely the same kernel statement. Interchanging the curves similarly gives $\ker {\alpha }^{\prime }=\varphi (E(\mathbb{Q}))$.

A prime not dividing $b$ can occur only with even valuation in an $x$-coordinate. If its valuation is positive, the factor $x^2+ax+b$ is a unit and $2v(y)=v(x)$; if negative, $x^2$ dominates that factor and $2v(y)=3v(x)$. Thus square classes in $\mathrm{im}\alpha$ have signed square-free representatives dividing $b$, giving finitely many candidates. To test a candidate $d$, put $x=d(u/v{)}^2$ and $y=duN/v^3$. The curve equation becomesA nontrivial rational solution, including the appropriate limiting points, is exactly a realization of the class. Clear denominators and arrange $\gcd (u,v)=1$ for congruence tests. Do the same for $E^{\prime }$ with $b^{\prime }$.

The factor in the two-isogeny rank formula can also be checked directly. Let $A=E(\mathbb{Q})$, $B=E^{\prime }(\mathbb{Q})$ and $t=\#A[2]$. The index of $\hat{\varphi }(B)$ in $A$ is $\#\mathrm{im}\alpha$, while the map from $B/\varphi (A)$ onto $\hat{\varphi }(B)/2A$ has kernelThe numerator has order two. Moreover $\ker \hat{\varphi }\cap \varphi (A)=\varphi (A[2])$, since $\hat{\varphi }\varphi P=2P$, and the latter has order $t/2$. Hence this quotient has order $4/t$, andThe Mordell-Weil theorem gives $[A:2A]=2^{r}t$, where $r$ is the rank. Cancelling $t$ yieldsIn practice, local congruence and real-solubility tests remove many candidate quartics, giving an upper bound, while rational point searches realize classes and give a lower bound. These often coincide. Local solubility alone is not a proof of a rational point: an everywhere locally soluble covering can still fail globally, so a remaining gap must be reported rather than treated as an exact rank.

For the curve at hand, $a=8$, $b=-7$, $b^{\prime }=92$, andThe first image is contained in $\{1,-1,7,-7\}$ and contains $1,-7$, the latter from the point $(0,0)$. To exclude $-1$, its quartic would require coprime $u,v$ withModulo $16$, an odd fourth power is $1$ and an even fourth power is $0$. If $u$ is odd and $v$ even the right side is $15$; if $u$ is even and $v$ odd it is $7$; if both are odd it is $14$. None is among the square residues $0,1,4,9$. Thus $-1$ is impossible. The image is a subgroup, so $7$ is also impossible: its product with the known class $-7$ would be $-1$. We obtain $\mathrm{im}\alpha =\{1,-7\}$.

For $E^{\prime }$, its quadratic factor is $X^2-16X+92=(X-8{)}^2+28>0$. Every nonzero rational $X$-coordinate is therefore positive. Prime support leaves only $1,2,23,46$ as possible classes; $1$ and $23=[92]$ are supplied by $O$ and the two-torsion point. The explicit point $(18,48)$ is on $E^{\prime }$ becauseIts class is $[18]=2$. Multiplying classes supplies $46$, so $\mathrm{im}{\alpha }^{\prime }=\{1,2,23,46\}$. The two-isogeny rank formula now provesAs an additional concrete point on the original curve, the dual isogeny sends $(18,48)$ to $(16/9,116/27)$. The descent calculation proves the rank; it does not by itself certify that this point generates the full free quotient.