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Write the matrix as $M(a,b,x)$. Direct multiplication and inversion giveThe identity is $M(0,0,0)$ and all determinants are one, so these formulas verify the subgroup criterion inside ${\mathrm{GL}}_3(\mathbb{R})$. This is the real Heisenberg group.

The map $M(a,b,x)\mapsto a$ is a surjective group homomorphism to the additive group of real numbers; its kernel is exactly $H$. Hence $H$ is normal and the first isomorphism theorem givesTwo matrices commute exactly when $a{b}^{\prime }=a^{\prime }b$. Requiring that relation for every $a^{\prime },b^{\prime }$ forces $a=b=0$, while $x$ is unrestricted. ThereforeFinally $M(a,b,x)\mapsto (a,b)$ is a surjective group homomorphism to the additive group ${\mathbb{R}}^2$ with precisely this kernel of a group homomorphism, giving $G/Z(G)\cong ({\mathbb{R}}^2,+)$. The central coordinate records noncommutativity: the group commutator is $M(0,0,a{b}^{\prime }-a^{\prime }b)$.

Write the Heisenberg group element as $M(x,y,z)$. Matrix multiplication and inversion giveAll matrices have determinant one. They contain $M(0,0,0)=I$, are closed under multiplication and inverses, and inherit associativity from matrix multiplication; hence they form a subgroup of $G{L}_3(\mathbb{R})$.

Two such matrices commute precisely when $x{z}^{\prime }=x^{\prime }z$. Requiring this for every $x^{\prime },z^{\prime }$ forces $x=z=0$. ConsequentlyThe map $\pi (M(x,y,z))=(x,z)$ is a surjective group homomorphism to the additive group ${\mathbb{R}}^2$, with kernel of a group homomorphism $Z(H)$. Equivalently, it gives the explicit quotient group isomorphismIt is well-defined because multiplying by a central element changes only $y$. It is injective because equal $(x,z)$ imply that the quotient of the two matrices is central, and surjectivity follows by taking $y=0$. This also proves the needed instance of the first isomorphism theorem directly.

The evaluation map $\phi :\mathbb{R}[x]\to \mathbb{C}$, $p\mapsto p(i)$, is a ring homomorphism and is onto because $a+bx$ maps to any $a+bi$. By polynomial division, writeThen $p(i)=0$ precisely when $a=b=0$. Thus $\ker \phi =(x^2+1)$, and the first isomorphism theorem givesSince the quotient ring is a field, $(x^2+1)$ is a maximal ideal.

Evaluation at a real point defines a surjective ring homomorphism $p\mapsto p(a,b)$ from $\mathbb{R}[x,y]$ to $\mathbb{R}$. Its kernel is $(x-a,y-b)$: subtracting the constant $p(a,b)$ expresses the remainder as a multiple of $x-a$ plus a multiple of $y-b$, by successive polynomial division. The first isomorphism theorem therefore identifies the quotient ring with the field $\mathbb{R}$, proving each $I_{a,b}$ is maximal.

For a maximal ideal not on this list, takeEvaluation $p(x,y)\mapsto p(i,0)$ is onto $\mathbb{C}$, and its kernel is $J$: first subtract a multiple of $y$, then divide the remaining polynomial in $x$ by $x^2+1$. Hence $\mathbb{R}[x,y]/J\cong \mathbb{C}$, a field. Finally, if $J=I_{a,b}$ for real $a,b$, the member $x^2+1\in J$ would vanish at $(a,b)$, requiring $a^2+1=0$, which is impossible over $\mathbb{R}$.

If $M=0$, take the empty filtration. Otherwise a nonzero quotient $Q=M/M_i$ is a finitely generated module. Because $R$ has the ascending chain condition on ideals, the nonempty collection of annihilators of nonzero elements of $Q$ has a maximal element. Part (b), applied to $Q$, supplies $q\ne 0$ with a prime ideal $P=\mathrm{Ann}(q)$.

The map $R\to Rq$, $r\mapsto rq$, is surjective with kernel $P$. The first isomorphism theorem gives $Rq\cong R/P$. Let $M_{i+1}$ be the inverse image of $Rq$ under $M\to Q$. ThenRepeat whenever the quotient remains nonzero. An infinite repetition would give a strictly ascending chain of submodules of the Noetherian module $M$, contradicting part (a). Thus the process stops after finitely many steps, necessarily at $M$.

This constructs the required prime filtration $0=M_0⊊\cdots ⊊M_l=M$, with each successive quotient $R/P_i$ for a prime $P_i$.

Let $G$ act by left multiplication on the set $G/H$ of $n$ left cosets. This gives a group action homomorphismIts kernelis a normal subgroup of $G$, and the first isomorphism theorem gives $G/K\cong \mathrm{im}\phi$. By Lagrange's theorem, $|G/K|$ divides $n!$. The image acts transitively on the $n$ cosets, so the orbit-stabilizer theorem shows that $n$ divides $|G/K|$; in particular,

Now suppose $G$ is nonabelian and simple. Since $H$ is proper, the coset action is nontrivial, so $K\ne G$. Simplicity gives $K=\{e\}$, and $\phi$ embeds $G$ into $S_n$. Composing with the sign homomorphism givesIts kernel is normal. A nontrivial map would embed the simple group $G$ into the abelian group of order two, which is impossible because $G$ is nonabelian. The sign is therefore always $+1$, so

A subgroup $N\le G$ is normal whenIts left and right cosets then agree, and multiplicationis well defined on the cosets. The identity is $N$, the inverse of $gN$ is $g^{-1}N$, and associativity descends from $G$, so these cosets form the quotient group $G/N$.

For a group homomorphism $\theta :G\to H$, its kernel and image areThe kernel is normal becausewhenever $k\in \ker \theta$. Conversely, if $K⊴G$, the quotient mapis a homomorphism with kernel $K$. The image of any homomorphism is closed under products and inverses, so it is a subgroup of $H$. Finally, the mapis well defined and bijective and preserves multiplication. This is the first isomorphism theorem.

DefineEuler's formula shows that $\Phi$ is a homomorphism, its image is the complex unit circle, and its kernel is $\mathbb{Z}$. The first isomorphism theorem therefore givesThe image of $\mathbb{Q}/\mathbb{Z}$ consists exactly of the roots of unity: if $t=p/q$, then $\Phi (t{)}^q=1$, while every element of finite order on the unit circle has an argument that is a rational multiple of $2\pi$.

The first isomorphism theorem says that for a homomorphism $\varphi :G\to H$,The map is well-defined because equal cosets differ by a kernel element; it is a surjective homomorphism onto the image, and injectivity follows because its kernel is the identity coset.

Here $G{L}_n(\mathbb{R})$ consists of invertible real $n\times n$ matrices and $S{L}_n(\mathbb{R})$ consists of those with determinant one. The latter is the kernel of the surjective determinant map to ${\mathbb{R}}^{\ast }$, so it is normal and

The given integral matrices form a group because products and inverses remain integral. The inverse formula shows that an integral inverse exists exactly when the determinant divides every cofactor; taking determinants shows more directly that $\det A\det A^{-1}=1$, hence $\det A=\pm 1$, and the adjugate formula proves the converse. The matrices $\left(\begin{array}{cc}1& m\\ 0& 1\end{array}\right)$ show infinitude.

Reduction modulo two is a homomorphism $G\to G{L}_2({\mathbb{F}}_2)$. Its kernel is exactly $H$, so $H$ is normal; its index is finite because the target has only six elements.

Let $G=\mathrm{Gal}(L/K)$. The lower ramification groups are $G_{-1}=G$ and, for $i\ge 0$,

If $L/K$ is totally ramified and ${\pi }_L$ is a uniformizer, then ${\mathcal{O}}_L={\mathcal{O}}_K[{\pi }_L]$. Factoring $F(\sigma {\pi }_L)-F({\pi }_L)$ by $\sigma {\pi }_L-{\pi }_L$ for $F\in {\mathcal{O}}_K[X]$ proves the uniformizer criterion for lower ramification groups

For $\sigma \in G_0$, defineThe inertia group $G_0$ acts trivially on $k_L$, so $\theta (\sigma \tau )=\theta (\sigma )\theta (\tau )$. Its kernel consists exactly of those $\sigma$ for which $\sigma ({\pi }_L)/{\pi }_L\equiv 1$ modulo the maximal ideal, namely $G_1$. The first isomorphism theorem therefore gives an injection