Proof: ($\Leftarrow$) We clearly have $a∈(b)=Rb$, and since $u^{-1}a=u^{-1}(ub)=b$, we have $b∈(a)=Ra$. Therefore, $Ra=(a) \subseteq (b)$ and $Rb=(b) \subseteq (a)$ so that $(a)=(b)$. ($⇒$) If either of $a$ or $b$ are zero, then both are zero, and the case clearly holds, so assume $a≠0≠b$. We must have $a=ub$ and $va=b$ for some $u,v∈R$. We can compare the two to obtain $va=vub=b$, so that $vub-b=(vu-1)b=0$, and now $vu-1=0$ implying $vu=uv=1$, i.e. $u$ is a unit as claimed.$~\square$
Tuesday, April 30, 2013
Principle Ideals in Integral Domains (7.4.8)
Dummit and Foote Abstract Algebra, section 7.4, exercise 8:
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Let $R$ be an integral domain with $a,b∈R$. Prove $(a)=(b)$ if and only if $a=ub$ for some unit $u$.
Proof: ($\Leftarrow$) We clearly have $a∈(b)=Rb$, and since $u^{-1}a=u^{-1}(ub)=b$, we have $b∈(a)=Ra$. Therefore, $Ra=(a) \subseteq (b)$ and $Rb=(b) \subseteq (a)$ so that $(a)=(b)$. ($⇒$) If either of $a$ or $b$ are zero, then both are zero, and the case clearly holds, so assume $a≠0≠b$. We must have $a=ub$ and $va=b$ for some $u,v∈R$. We can compare the two to obtain $va=vub=b$, so that $vub-b=(vu-1)b=0$, and now $vu-1=0$ implying $vu=uv=1$, i.e. $u$ is a unit as claimed.$~\square$
Proof: ($\Leftarrow$) We clearly have $a∈(b)=Rb$, and since $u^{-1}a=u^{-1}(ub)=b$, we have $b∈(a)=Ra$. Therefore, $Ra=(a) \subseteq (b)$ and $Rb=(b) \subseteq (a)$ so that $(a)=(b)$. ($⇒$) If either of $a$ or $b$ are zero, then both are zero, and the case clearly holds, so assume $a≠0≠b$. We must have $a=ub$ and $va=b$ for some $u,v∈R$. We can compare the two to obtain $va=vub=b$, so that $vub-b=(vu-1)b=0$, and now $vu-1=0$ implying $vu=uv=1$, i.e. $u$ is a unit as claimed.$~\square$
Nilpotency and the Augmentation Ideal (7.4.3)
Dummit and Foote Abstract Algebra, section 7.4, exercise 3:
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(a) Let $G$ be an abelian group of order $p^n$. Prove $Aug(\mathbb{F}_pG)=\mathfrak{N}(\mathbb{F}_pG)$.
(b) Let $G=\{g_1,...,g_n\}$ be a group and assume $R$ is commutative. Prove $r∈Aug(RG)⇒r(g_1+...+g_n)=0$.
Proof: (a) ($\subseteq$) By the preceding exercise, $Aug(\mathbb{F}_pG)$ is generated by $\{g-1~|~g∈G\}$, so for arbitrary $x∈Aug(\mathbb{F}_pG)$ write $x=\sum_{g∈G}a_g(g-1)$ for some $a_g∈\mathbb{F}_pG$ for all $g∈G$. Prove $(g-1)^{p^n}=0$ and is thus nilpotent for any $g∈G$, so that by the idealness of $\mathfrak{N}(\mathbb{F}_pG)$ we have $x$ is nilpotent. By the binomial theorem for commutative rings, since $p^n$ divides $p^n \choose k$$ = \dfrac{p^n!}{k!(n-k)!}$ when $0 < k < p^n$, we can observe $(g-1)^{p^n}=g^{p^n}-1=1-1=0$. ($\supseteq$) Since $\mathbb{F}_pG/Aug(\mathbb{F}_pG)≅\mathbb{F}_p$ is a field, we have $Aug(\mathbb{F}_pG)$ is a maximal ideal, so that $\mathfrak{N}(\mathbb{F}_pG)∈\{Aug(\mathbb{F}_pG),\mathbb{F}_pG\}$, and is evidently not the latter as $1∈\mathbb{F}_pG$ is not nilpotent.$~\square$
(b) Once again write $r=r'(g-1)$ for some $r'∈RG$, and recalling that the action of multiplication on a group by one of its elements is a permutation of that group's elements, we observe $$r(g_1+...+g_n)=r'(g-1)(g_1+...+g_n)=r'(g(g_1+...+g_n)-(g_1+...+g_n))=$$$$r'((g_1+...+g_n)-(g_1+...+g_n))=r'0=0~\square$$
(b) Let $G=\{g_1,...,g_n\}$ be a group and assume $R$ is commutative. Prove $r∈Aug(RG)⇒r(g_1+...+g_n)=0$.
Proof: (a) ($\subseteq$) By the preceding exercise, $Aug(\mathbb{F}_pG)$ is generated by $\{g-1~|~g∈G\}$, so for arbitrary $x∈Aug(\mathbb{F}_pG)$ write $x=\sum_{g∈G}a_g(g-1)$ for some $a_g∈\mathbb{F}_pG$ for all $g∈G$. Prove $(g-1)^{p^n}=0$ and is thus nilpotent for any $g∈G$, so that by the idealness of $\mathfrak{N}(\mathbb{F}_pG)$ we have $x$ is nilpotent. By the binomial theorem for commutative rings, since $p^n$ divides $p^n \choose k$$ = \dfrac{p^n!}{k!(n-k)!}$ when $0 < k < p^n$, we can observe $(g-1)^{p^n}=g^{p^n}-1=1-1=0$. ($\supseteq$) Since $\mathbb{F}_pG/Aug(\mathbb{F}_pG)≅\mathbb{F}_p$ is a field, we have $Aug(\mathbb{F}_pG)$ is a maximal ideal, so that $\mathfrak{N}(\mathbb{F}_pG)∈\{Aug(\mathbb{F}_pG),\mathbb{F}_pG\}$, and is evidently not the latter as $1∈\mathbb{F}_pG$ is not nilpotent.$~\square$
(b) Once again write $r=r'(g-1)$ for some $r'∈RG$, and recalling that the action of multiplication on a group by one of its elements is a permutation of that group's elements, we observe $$r(g_1+...+g_n)=r'(g-1)(g_1+...+g_n)=r'(g(g_1+...+g_n)-(g_1+...+g_n))=$$$$r'((g_1+...+g_n)-(g_1+...+g_n))=r'0=0~\square$$
Friday, April 26, 2013
Nonisomorphicity of Z[x] and Q[x] (7.3.2)
Dummit and Foote Abstract Algebra, section 7.3, exercise 2:
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Prove $\mathbb{Z}[x] \not ≅ \mathbb{Q}[x]$.
Proof: Lemma 1: If $φ:R→S$ is a surjective ring homomorphism and $R$ is generated by a subset $X$ (i.e. every element of $R$ is writable as a sum of products with subtraction [or a series of operations] of elements from $X$), then $S$ is generated by $φ(X)$. Proof: Let $s$ be an arbitrary element of $S$. For some $r∈R$, we have $s=φ(r)$ and since $r$ is a series of operations of elements from $X$, due to the homomorphism's nature we have $s$ is a series of operations of elements from $φ(X)$.$~\square$
We can see that $\mathbb{Z}[x]$ is generated by $\{x,1\}$, since the latter generates $\mathbb{Z}$ (all potential coefficients) and the former generates $x^n$ for all $n∈\mathbb{Z}^+$. This implies $\mathbb{Q}[x]$ is generated by two elements, and in particular $\mathbb{Q}$ is generated by two elements (since we have a surjective homomorphism $φ:\mathbb{Q}[x] → \mathbb{Q}$ by $p(x) \mapsto p(0)$). Letting $\dfrac{a}{b}$ and $\dfrac{c}{d}$ denote these generators, we have $ad \cdot \dfrac{1}{bd} = \dfrac{a}{b}$ and $bc \cdot \dfrac{1}{bd}=\dfrac{c}{d}$, so that in fact $\mathbb{Q}$ is generated by one element; further refer to it as $\dfrac{1}{z}$. Letting $p \not \mid z$ be prime, we must have $\dfrac{1}{p}$ is generated by a series of operations of $\dfrac{1}{z}$ despite the fact that every such operation can be condensed into some fraction $\dfrac{y}{z^n}$, which cannot simplify to $\dfrac{1}{p}$.$~\square$
Proof: Lemma 1: If $φ:R→S$ is a surjective ring homomorphism and $R$ is generated by a subset $X$ (i.e. every element of $R$ is writable as a sum of products with subtraction [or a series of operations] of elements from $X$), then $S$ is generated by $φ(X)$. Proof: Let $s$ be an arbitrary element of $S$. For some $r∈R$, we have $s=φ(r)$ and since $r$ is a series of operations of elements from $X$, due to the homomorphism's nature we have $s$ is a series of operations of elements from $φ(X)$.$~\square$
We can see that $\mathbb{Z}[x]$ is generated by $\{x,1\}$, since the latter generates $\mathbb{Z}$ (all potential coefficients) and the former generates $x^n$ for all $n∈\mathbb{Z}^+$. This implies $\mathbb{Q}[x]$ is generated by two elements, and in particular $\mathbb{Q}$ is generated by two elements (since we have a surjective homomorphism $φ:\mathbb{Q}[x] → \mathbb{Q}$ by $p(x) \mapsto p(0)$). Letting $\dfrac{a}{b}$ and $\dfrac{c}{d}$ denote these generators, we have $ad \cdot \dfrac{1}{bd} = \dfrac{a}{b}$ and $bc \cdot \dfrac{1}{bd}=\dfrac{c}{d}$, so that in fact $\mathbb{Q}$ is generated by one element; further refer to it as $\dfrac{1}{z}$. Letting $p \not \mid z$ be prime, we must have $\dfrac{1}{p}$ is generated by a series of operations of $\dfrac{1}{z}$ despite the fact that every such operation can be condensed into some fraction $\dfrac{y}{z^n}$, which cannot simplify to $\dfrac{1}{p}$.$~\square$
Thursday, April 25, 2013
A Ring of Homomorphisms (7.1.30)
Dummit and Foote Abstract Algebra, section 7.1, exercise 30:
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Let $A=\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} ...$ be the direct product of copies of $\mathbb{Z}$ indexed by positive integers and let $R$ be the ring of homomorphisms from $A$ to itself. Let $φ$ be the element of $R$ defined by $φ(a_1,a_2,a_3,...)=(a_2,a_3,...)$. Let $ψ$ be the element of $R$ defined by $ψ(a_1,a_2,a_3,...)=(0,a_1,a_2,a_3,...)$.
(a) Prove $φψ=1≠ψφ$.
(b) Exhibit infinitely many right inverses for $φ$.
(c) Find a nonzero $\pi∈R$ such that $φ\pi = 0 ≠ \piφ$.
(d) Prove $\nexists \lambda∈R~(\lambdaφ=0)$.
Proof: (a)$$(φψ)(a_1,a_2,a_3,...)=φ(ψ(a_1,a_2,a_3,...))=φ(0,a_1,a_2,...)=(a_1,a_2,a_3,...)$$(b) Let $\pi$ be the homomorphism defined in part (c). We claim $\pi^n+ψ$ is a right inverse for any $n∈\mathbb{Z}^+$. We have $φ(\pi^n+ψ)=φ\pi\pi^{n-1}+φψ=φψ=1$, so now it suffices to prove $\pi^n+ψ≠\pi^m+ψ$ for any $m≠n$: $$\pi^n+ψ=\pi^m+ψ⇒\pi^n=\pi^m⇒$$$$\pi^n(1,0,...)=\pi^m(1,0,...)⇒(2^n,0,...)=(2^m,0,...)⇒n=m$$(c) Let $\pi$ be defined by $\pi(a_1,a_2,a_3,...)=(2a_1,0,0,...)$. Prove it is a homomorphism: $\pi((a_1,a_2,...)+(b_1,b_2,...))=\pi((a_1+b_1,a_2+b_2,...)=(2(a_1+b_1),0,...)=$$(2a_1,0,...)+(2b_1,0,...)=\pi(a_1,a_2,...)+\pi(b_1,b_2,...)$. We have $(φ\pi)(a_1,a_2,...)=φ(\pi(a_1,a_2,))=φ(2a_1,0,...)=(0,0,...)$, so that $φ\pi=0$. Now, $(\piφ)(0,1,0,0,...)=(2,0,0,...)$, so that $\piφ≠0$.
(d) By exercise 28(b), since $φ$ has $ψ$ a right inverse, $φ$ is not a right zero divisor.$~\square$
(a) Prove $φψ=1≠ψφ$.
(b) Exhibit infinitely many right inverses for $φ$.
(c) Find a nonzero $\pi∈R$ such that $φ\pi = 0 ≠ \piφ$.
(d) Prove $\nexists \lambda∈R~(\lambdaφ=0)$.
Proof: (a)$$(φψ)(a_1,a_2,a_3,...)=φ(ψ(a_1,a_2,a_3,...))=φ(0,a_1,a_2,...)=(a_1,a_2,a_3,...)$$(b) Let $\pi$ be the homomorphism defined in part (c). We claim $\pi^n+ψ$ is a right inverse for any $n∈\mathbb{Z}^+$. We have $φ(\pi^n+ψ)=φ\pi\pi^{n-1}+φψ=φψ=1$, so now it suffices to prove $\pi^n+ψ≠\pi^m+ψ$ for any $m≠n$: $$\pi^n+ψ=\pi^m+ψ⇒\pi^n=\pi^m⇒$$$$\pi^n(1,0,...)=\pi^m(1,0,...)⇒(2^n,0,...)=(2^m,0,...)⇒n=m$$(c) Let $\pi$ be defined by $\pi(a_1,a_2,a_3,...)=(2a_1,0,0,...)$. Prove it is a homomorphism: $\pi((a_1,a_2,...)+(b_1,b_2,...))=\pi((a_1+b_1,a_2+b_2,...)=(2(a_1+b_1),0,...)=$$(2a_1,0,...)+(2b_1,0,...)=\pi(a_1,a_2,...)+\pi(b_1,b_2,...)$. We have $(φ\pi)(a_1,a_2,...)=φ(\pi(a_1,a_2,))=φ(2a_1,0,...)=(0,0,...)$, so that $φ\pi=0$. Now, $(\piφ)(0,1,0,0,...)=(2,0,0,...)$, so that $\piφ≠0$.
(d) By exercise 28(b), since $φ$ has $ψ$ a right inverse, $φ$ is not a right zero divisor.$~\square$
Order of Conductor f (7.1.23)
Dummit and Foote Abstract Algebra, section 7.1, exercise 23:
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Define $\mathcal{O}_f=\mathbb{Z}[f\omega]=\{a+b\omega~|~a,b∈\mathbb{Z}\}$. Letting $[R]$ denote the abelian group in the ring $R$, prove:$$R = \mathcal{O}_f ⇔ R~\text{subring of}~\mathcal{O} \land 1∈R \land |[\mathcal{O}] : [R]|=f$$
Proof: Lemma 1: For $H≤G$ groups and $φ$ an isomorphism, we have $|G:H|=|φ(G):φ(H)|$. Proof: Let $n_1,...$ be a complete set of coset-distinct representatives for $H$ in $G$. We claim that their images are a complete set of coset-distinct representatives of $φ(H)$ in $φ(G)$. For any $g∈G$, we have $g=n_ih$ for some index $i$ and $h∈H$. Therefore, we have $φ(g)=φ(n_ih)=φ(n_i)φ(h)$. Assuming $φ(n_i)$ and $φ(n_j)$ are in the same coset of $φ(H)$, we have $φ(n_i)=φ(n_j)φ(h)$$=φ(n_jh)$, so that $n_i=n_jh$ and $n_i$ and $n_j$ are in the same coset of $H$, a contradiction. Thus, there are just as many coset representatives of $φ(H)$ in $φ(G)$ as $H$ in $G$.$~\square$
($⇒$) (1) All that needs to be done is to prove closure under subtraction and multiplication. We have $(a+bf\omega)-(c+df\omega)=(a-c)+(b-d)f\omega$. As well, $\omega^2=D$ or $\dfrac{D-1}{4}+\dfrac{1+\sqrt{D}}{2}$ depending on $D$ modulo 4, so$$(a+bf\omega)(c+df\omega)=(ac)+(bc+ad)f\omega+bdf^2\omega^2$$is evidently within the ring as well. (2) Clearly $1∈\mathcal{O}_f$; set $a=1$ and $b=0$. (3) Construct a mapping $φ:[\mathcal{O}]→\mathbb{Z} \times \mathbb{Z}$ by $a+b\omega \mapsto (a,b)$. This is well defined by unique $a,b$ form of the argument, clearly surjective and injective, and$$φ((a+b\omega)+(c+d\omega))=φ((a+c)+(b+d)\omega)=(a+c,b+d)$$$$=(a,b)+(c,d)=φ(a+b\omega)+φ(c+d\omega)$$so that $φ$ is homomorphic and now an isomorphism. So $[\mathcal{O}]≅\mathbb{Z} \times \mathbb{Z}$ and by definition we can tell $[\mathcal{O}_f] ≅ φ([\mathcal{O}_f])=\mathbb{Z} \times f\mathbb{Z}$. By 5.1.14, we have $(\mathbb{Z} \times \mathbb{Z})/(\mathbb{Z} \times f\mathbb{Z})≅$$(\mathbb{Z}/\mathbb{Z})\times(\mathbb{Z}/f\mathbb{Z})≅\mathbb{Z}/f\mathbb{Z}$ is a group of order $f$, so that $f=|\mathbb{Z} \times \mathbb{Z} : \mathbb{Z} \times f\mathbb{Z}|=|φ([\mathcal{O}]):φ([\mathcal{O}_f])|=|[\mathcal{O}]:[\mathcal{O}_f]|$.
Lemma 2: Any nontrivial subgroup of $\mathbb{Z}$ is of the form $n\mathbb{Z}$ for some $n∈\mathbb{Z}^+$. Proof: Let distinct $a,b$ be chosen with $n=|a-b|$ minimal; we thus have $n\mathbb{Z}$ is contained within the subgroup. Assume there is an element $x$ not within $n\mathbb{Z}$: We have $x=qn+r$ with $0 < r < n$, and since $qn∈n\mathbb{Z}$, we have $x-qn < n=|a-b|$, a contradiction. So $n\mathbb{Z}$ is precisely the subgroup in question.$~\square$
($\Leftarrow$) $R$'s identity must be $1$, as $n(a+bf\omega)≠a+bf\omega$ for any $n≠1$ and $a≠0≠b$. Therefore, we have $1\mathbb{Z} \times 0=\mathbb{Z} \times 0≤φ([R])$. For any $(a,b)∈φ([R])$, we also have $(a,0)∈φ([R])$, so that $(0,b)∈φ([R])$, implying the second factor of $φ([R])$ is a subgroup of $\mathbb{Z}$ and therefore of the form $n\mathbb{Z}$. Since $| \mathbb{Z} \times \mathbb{Z}:\mathbb{Z} \times n\mathbb{Z}|=n$, we have $n=f$ and now $φ([R])=\mathbb{Z} \times f\mathbb{Z}=φ([\mathcal{O}_f])$ and now $[R]=[\mathcal{O}_f]$. Since the specifications of multiplication are predefined in $\mathbb{C}$, we have $R=\mathcal{O}_f$.$~\square$
($⇒$) (1) All that needs to be done is to prove closure under subtraction and multiplication. We have $(a+bf\omega)-(c+df\omega)=(a-c)+(b-d)f\omega$. As well, $\omega^2=D$ or $\dfrac{D-1}{4}+\dfrac{1+\sqrt{D}}{2}$ depending on $D$ modulo 4, so$$(a+bf\omega)(c+df\omega)=(ac)+(bc+ad)f\omega+bdf^2\omega^2$$is evidently within the ring as well. (2) Clearly $1∈\mathcal{O}_f$; set $a=1$ and $b=0$. (3) Construct a mapping $φ:[\mathcal{O}]→\mathbb{Z} \times \mathbb{Z}$ by $a+b\omega \mapsto (a,b)$. This is well defined by unique $a,b$ form of the argument, clearly surjective and injective, and$$φ((a+b\omega)+(c+d\omega))=φ((a+c)+(b+d)\omega)=(a+c,b+d)$$$$=(a,b)+(c,d)=φ(a+b\omega)+φ(c+d\omega)$$so that $φ$ is homomorphic and now an isomorphism. So $[\mathcal{O}]≅\mathbb{Z} \times \mathbb{Z}$ and by definition we can tell $[\mathcal{O}_f] ≅ φ([\mathcal{O}_f])=\mathbb{Z} \times f\mathbb{Z}$. By 5.1.14, we have $(\mathbb{Z} \times \mathbb{Z})/(\mathbb{Z} \times f\mathbb{Z})≅$$(\mathbb{Z}/\mathbb{Z})\times(\mathbb{Z}/f\mathbb{Z})≅\mathbb{Z}/f\mathbb{Z}$ is a group of order $f$, so that $f=|\mathbb{Z} \times \mathbb{Z} : \mathbb{Z} \times f\mathbb{Z}|=|φ([\mathcal{O}]):φ([\mathcal{O}_f])|=|[\mathcal{O}]:[\mathcal{O}_f]|$.
Lemma 2: Any nontrivial subgroup of $\mathbb{Z}$ is of the form $n\mathbb{Z}$ for some $n∈\mathbb{Z}^+$. Proof: Let distinct $a,b$ be chosen with $n=|a-b|$ minimal; we thus have $n\mathbb{Z}$ is contained within the subgroup. Assume there is an element $x$ not within $n\mathbb{Z}$: We have $x=qn+r$ with $0 < r < n$, and since $qn∈n\mathbb{Z}$, we have $x-qn < n=|a-b|$, a contradiction. So $n\mathbb{Z}$ is precisely the subgroup in question.$~\square$
($\Leftarrow$) $R$'s identity must be $1$, as $n(a+bf\omega)≠a+bf\omega$ for any $n≠1$ and $a≠0≠b$. Therefore, we have $1\mathbb{Z} \times 0=\mathbb{Z} \times 0≤φ([R])$. For any $(a,b)∈φ([R])$, we also have $(a,0)∈φ([R])$, so that $(0,b)∈φ([R])$, implying the second factor of $φ([R])$ is a subgroup of $\mathbb{Z}$ and therefore of the form $n\mathbb{Z}$. Since $| \mathbb{Z} \times \mathbb{Z}:\mathbb{Z} \times n\mathbb{Z}|=n$, we have $n=f$ and now $φ([R])=\mathbb{Z} \times f\mathbb{Z}=φ([\mathcal{O}_f])$ and now $[R]=[\mathcal{O}_f]$. Since the specifications of multiplication are predefined in $\mathbb{C}$, we have $R=\mathcal{O}_f$.$~\square$
Wednesday, April 24, 2013
Commutativity of Boolean Rings (7.1.15-16)
Dummit and Foote Abstract Algebra, section 7.1, exercises 15-16:
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15. A ring $R$ is a Boolean ring if $a^2=a$ for all $a∈R$. Prove that every Boolean ring is commutative.
16. Prove that the only Boolean integral domain is $\mathbb{Z}/2\mathbb{Z}$.
Proof: (15) Take any $a,b∈R$. Note that we necessarily have $(-x)^2=-x$ for any $x∈R$ as well as $(-x)^2=(-x)(-x)=x^2=x$, so that $-x=x$. Now, notice that $(a+b)^2=a+b$ by the Boolean property. By distributing, we have $(a+b)^2=a^2+ab+ba+b^2=a+ab+ba+b$. Comparing the two, we have $ab+ba=0$, so that $ab=-ba=ba$.$~\square$
(16) For any $a$ in a Boolean integral domain, we have $a(a-1)=a^2-a=a-a=0$, so that either $a=0$ or $a-1=0$ implying $a=1$.$~\square$
16. Prove that the only Boolean integral domain is $\mathbb{Z}/2\mathbb{Z}$.
Proof: (15) Take any $a,b∈R$. Note that we necessarily have $(-x)^2=-x$ for any $x∈R$ as well as $(-x)^2=(-x)(-x)=x^2=x$, so that $-x=x$. Now, notice that $(a+b)^2=a+b$ by the Boolean property. By distributing, we have $(a+b)^2=a^2+ab+ba+b^2=a+ab+ba+b$. Comparing the two, we have $ab+ba=0$, so that $ab=-ba=ba$.$~\square$
(16) For any $a$ in a Boolean integral domain, we have $a(a-1)=a^2-a=a-a=0$, so that either $a=0$ or $a-1=0$ implying $a=1$.$~\square$
Tuesday, April 23, 2013
Free Groups and Nilpotency/Solvability (6.3.12)
Dummit and Foote Abstract Algebra, section 6.3, exercise 12:
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Let $S$ be a set and let $c$ be a positive integer. Formulate the notion of a free nilpotent group on $S$ of nilpotence class $c$ and prove it has the appropriate universal property with respect to nilpotent groups of class $≤c$.
Proof: Let $N$ be any group of nilpotence class $≤c$ and let $ψ: S → N$ be a set map with $G = \langle~\text{img}~ψ~\rangle$. We have a unique homomorphism $φ : F(S) → N$ (fixing $S$) so that $F(S)/\text{ker}~φ ≅ G$. We prove that there is a unique homomorphism $\Phi : F(S)/F(S)^c → N$ such that $\Phi\mid_{\pi(S)} = ψ$ (where $\pi$ is the natural homomorphism from $F(S)$ to $F(S)/F(S)^c$). Assume $F(S)^c \not ≤ \text{ker}~φ$; we then have a contradiction: $$G^c≅(F(S)/\text{ker}~φ)^c=F(S)^c/\text{ker}~φ≠1$$Therefore there is the desired homomorphism afforded by:$$(F(S)/F(S)^c)/(\text{ker}~φ/F(S)^c)≅F(S)/\text{ker}~φ≅G$$Assume $\Phi_1 ≠ \Phi_2$ are two homomorphisms from $F(S)/F(S)^c$ to $N$ fixing $\pi(S)$; then $\Phi_1 \circ \pi ≠ \Phi_2 \circ \pi$ are two homomorphisms from $F(S)$ to $G$ fixing $S$, a contradiction. We thus have $\text{ker}~φ$ factors through $F(S)^c$, and the following diagram commutes:
Note that this theorem can be paralleled to produce a similar result regarding a free solvable group on a set $S$.
Proof: Let $N$ be any group of nilpotence class $≤c$ and let $ψ: S → N$ be a set map with $G = \langle~\text{img}~ψ~\rangle$. We have a unique homomorphism $φ : F(S) → N$ (fixing $S$) so that $F(S)/\text{ker}~φ ≅ G$. We prove that there is a unique homomorphism $\Phi : F(S)/F(S)^c → N$ such that $\Phi\mid_{\pi(S)} = ψ$ (where $\pi$ is the natural homomorphism from $F(S)$ to $F(S)/F(S)^c$). Assume $F(S)^c \not ≤ \text{ker}~φ$; we then have a contradiction: $$G^c≅(F(S)/\text{ker}~φ)^c=F(S)^c/\text{ker}~φ≠1$$Therefore there is the desired homomorphism afforded by:$$(F(S)/F(S)^c)/(\text{ker}~φ/F(S)^c)≅F(S)/\text{ker}~φ≅G$$Assume $\Phi_1 ≠ \Phi_2$ are two homomorphisms from $F(S)/F(S)^c$ to $N$ fixing $\pi(S)$; then $\Phi_1 \circ \pi ≠ \Phi_2 \circ \pi$ are two homomorphisms from $F(S)$ to $G$ fixing $S$, a contradiction. We thus have $\text{ker}~φ$ factors through $F(S)^c$, and the following diagram commutes:
Note that this theorem can be paralleled to produce a similar result regarding a free solvable group on a set $S$.
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