MTH3003 Lecture 11

Last time we built up the machinery of quotients $G/N$ and saw how kernels of homomorphisms naturally give rise to factor groups. In this lecture we make that story precise via the First isomorphism theorem, then add two further structural results (the Second isomorphism theorem and Third isomorphism theorem) that let us shuffle subgroups and quotients around safely. We finish by introducing cycle shape in ${\mathrm{S}}_n$, setting up the signature map and the alternating group for the next lecture.

The First Isomorphism Theorem

We start with the fundamental relationship between a homomorphism, its kernel, and its image.

First isomorphism theorem

Let group homomorphism $\phi :G\to H$ have kernel $K=\ker \phi$ and image $I=\mathrm{Im}\phi$. If $K⊴G$ then the quotient group $G/K$ is a group and there is an isomorphism $\theta :G/K\to I$ given by $\theta (gK)=\phi (g)$. In particular, we obtain a canonical group isomorphism

$$\boxed{G/K\cong \mathrm{Im}\phi }.$$

We already know that if $K=\ker \phi$ then $K⊴G$, so $G/K$ exists as a group of cosets. The real work is to show that the formula $\theta (gK)=\phi (g)$ does not depend on the choice of representative $g$ and that $\theta$ is an isomorphism onto the image.

Proof Structure and “well-defined”

1. Well-definedness. Suppose $g_{1}K=g_{2}K$. Then $g_1^{-1}{g}_2\in K$, so $\phi (g_1^{-1}{g}_2)=e_H$. By the homomorphism property,

\varphi(g_{1}^{-1}g_{2}) = \varphi(g_{1})^{-1}\varphi(g_{2}) = e_{H} \implies \varphi(g_{1}) = \varphi(g_{2}).

Hence $\theta(g_{1}K) = \theta(g_{2}K)$, so $\theta$ is a genuine function on cosets. 2. **Homomorphism property.** For $g_{1}, g_{2} \in G$ we have

\theta\bigl((g_{1}K)(g_{2}K)\bigr) = \theta((g_{1}g_{2})K) = \varphi(g_{1}g_{2}) = \varphi(g_{1})\varphi(g_{2}) = \theta(g_{1}K),\theta(g_{2}K).

3. **Surjectivity onto $I$.** Every element of $I$ has the form $\varphi(g)$ for some $g \in G$, and $\varphi(g) = \theta(gK)$, so $\theta$ is onto $I$. 4. **Injectivity.** If $\theta(g_{1}K) = \theta(g_{2}K)$ then $\varphi(g_{1}) = \varphi(g_{2})$, so $\varphi(g_{1}^{-1}g_{2}) = e_{H}$ and hence $g_{1}^{-1}g_{2} \in K$. This is equivalent to $g_{1}K = g_{2}K$. Putting this together, $\theta$ is a bijective homomorphism $G/K \to I$, hence an isomorphism. > [!note] Canonical homomorphism > > Given $N \trianglelefteq G$, the map $\pi \colon G \to G/N$ defined by $\pi(g) = gN$ is a surjective homomorphism with $\ker \pi = N$. > It is often called the **canonical projection or canonical homomorphism**. From the first isomorphism theorem, any homomorphism $\varphi \colon G \to H$ with kernel $K$ factors through the quotient $G/K$ as

G \xrightarrow{\pi} G/K \xrightarrow[\cong]{\theta} \operatorname{Im}\varphi \leq H,

where $\pi$ is the canonical projection and $\theta$ is an isomorphism. ### Example: $\mathbb{Z}/n\mathbb{Z} \cong \mathbb{Z}_{n}$ Consider the homomorphism $\varphi \colon \mathbb{Z} \to \mathbb{Z}_{n}$ defined by

\varphi(m) = [m]*{n}.

We know: - $\ker \varphi = n\mathbb{Z}$, the set of all multiples of $n$. - $\operatorname{Im}\varphi = \mathbb{Z}_{n}$ (the map is surjective). By the first isomorphism theorem we obtain

\boxed{,\mathbb{Z}/n\mathbb{Z} ;\cong; \mathbb{Z}*{n},}.

This recovers our earlier explicit isomorphism (for example when $n = 3$) in a single conceptual statement valid for all $n \in \mathbb{N}$. --- ## The Second Isomorphism Theorem The second isomorphism theorem shows how a subgroup $H$ and a normal subgroup $N$ combine inside $G$. > [!important] Second isomorphism theorem > > Let $G$ be a group with $H \leq G$ and $N \trianglelefteq G$. Then: > - The product $HN = \{hn \colon h \in H, n \in N\}$ is a subgroup of $G$. > - The intersection $H \cap N$ is a normal subgroup of $H$. > - There is an isomorphism >

\boxed{,H/(H \cap N) ;\cong; (HN)/N,}.

Proof via the First Isomorphism Theorem

Define $\phi :H\to (HN)/N$ by

$$\phi (h)=hN.$$

1. Homomorphism. For $h_1,h_2\in H$ we have

$$\phi (h_1{h}_2)=(h_1{h}_2)N=h_1{h}_{2}N=h_{1}N\cdot h_{2}N=\phi (h_1)\phi (h_2),$$

because multiplication of cosets in $(HN)/N$ is inherited from $G$ and $N⊴G$.

1. Kernel. The kernel is

$$\ker \phi =\{h\in H:hN=N\}=\{h\in H:h\in N\}=H\cap N.$$

1. Image. Any element of $(HN)/N$ has the form $hnN=hN$ with $h\in H$ and $n\in N$, so

$$\mathrm{Im}\phi =(HN)/N.$$

By the first isomorphism theorem,

$$H/(H\cap N)\cong (HN)/N.$$

This is extremely useful when you want to compare the “copy of $H$ modulo its intersection with $N$” with the “chunk of $G$ generated by $H$ and $N$, modulo $N$“.

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The Third Isomorphism Theorem (“Fool’s Cancellation”)

The third isomorphism theorem formalises what looks like cancellation in a tower of quotients. The warning is that you can only “cancel” when the relevant subgroups are normal in the right groups.

Third isomorphism theorem

Let $G$ be a group and let $N⊴G$. Suppose $M$ is another normal subgroup with

$$>N\le M\le G,M⊴G.>$$

Then:

• The subgroup $M/N$ is normal in $G/N$.
• There is an isomorphism

$$\boxed{(G/N)/(M/N)\cong G/M}.$$

People sometimes write this heuristically as

$$(G/N)/(M/N)\cong G/M\text{"so we can cancel /N".}$$

The whole point of the theorem is that this is not true in general unless the normality conditions are satisfied.

Why "Fool's Cancellation"?

You cannot cancel arbitrary subgroups the way you cancel numbers. The theorem only works when $N⊴G$ and $M⊴G$ with $N\le M$. If you try to imitate this when normality fails, the quotient groups may not even make sense.

Proof via the First Isomorphism Theorem

Define $\phi :G/N\to G/M$ by

$$\phi (gN)=gM.$$

We must check that this is well-defined and a homomorphism, then identify its kernel and image.

1. Well-defined homomorphism. If $g_{1}N=g_{2}N$ then $g_1^{-1}{g}_2\in N\le M$, so $g_{1}M=g_{2}M$ and hence $\phi$ is well-defined. For $g_1,g_2\in G$ we have

$$\phi (g_{1}N\cdot g_{2}N)=\phi ((g_1{g}_2)N)=(g_1{g}_2)M=g_{1}M\cdot g_{2}M=\phi (g_{1}N)\phi (g_{2}N),$$

so $\phi$ is a homomorphism.

1. Kernel. The kernel is

$$\ker \phi =\{gN:gM=M\}=\{gN:g\in M\}=M/N.$$

This shows $M/N$ is normal in $G/N$.

1. Image. For any $g\in G$, $gM$ lies in $G/M$, and every coset of $M$ arises like this. Hence

$$\mathrm{Im}\phi =G/M.$$

By the first isomorphism theorem,

$$(G/N)/(M/N)\cong G/M.$$

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Cycle Shape in ${\mathrm{S}}_n$

We now move to a new topic: the cycle shape of permutations in the symmetric group ${\mathrm{S}}_n$. This is a way to record how long the disjoint cycles of a permutation are, ignoring which actual points are moved.

Definition: cycle shape

Let $g\in {\mathrm{S}}_n$ be a permutation, $g\ne e$. Write $g$ as a product of disjoint cycles

$$>g=c_1{c}_2\cdots c_m,>$$

where each cycle $c_i$ has length $r_i$ and $r_1\ge r_2\ge \cdots \ge r_m$. The cycle shape of $g$ is the tuple

$$>\boxed{(r_1,r_2,\dots ,r_m)}.>$$

The cycle shape of the identity permutation is usually taken to be $\varnothing$.

This is an invariant of a permutation up to relabelling of the underlying set and reordering of disjoint cycles.

Examples of cycle shape

1. In ${\mathrm{S}}_{12}$, consider

$$g=(135729)(48)(1061211).$$

The cycles have lengths $6$, $2$, and $4$, so the cycle shape is $(6,4,2)$ after reordering. 2. In ${\mathrm{S}}_{10}$, let

$$h=(12)(34)(78)(910).$$

Here $h$ is a product of four disjoint transpositions, so the cycle shape is $(2,2,2,2)$.

The key point is that cycle shape forgets the actual labels and remembers only the multiset of cycle lengths.

All Cycle Shapes in ${\mathrm{S}}_4$

In ${\mathrm{S}}_4$, every permutation is a product of disjoint cycles acting on $\{1,2,3,4\}$, and each element lies in exactly one cycle (possibly a $1$-cycle). The possible cycle shapes and associated permutations can be listed exhaustively.

We have:

Cycle shape	Elements with this shape in ${\mathrm{S}}_4$	Number of elements
$\varnothing$	$e$	$1$
$(2)$	$(12)$, $(13)$, $(14)$, $(23)$, $(24)$, $(34)$	$6$
$(2,2)$	$(12)(34)$, $(13)(24)$, $(14)(23)$	$3$
$(3)$	$(123)$, $(132)$, $(124)$, $(142)$, $(134)$, $(143)$, $(234)$, $(243)$	$8$
$(4)$	$(1234)$, $(1243)$, $(1324)$, $(1342)$, $(1423)$, $(1432)$	$6$

We have definitely listed all elements, because

$$1+6+3+8+6=24=4!=|{\mathrm{S}}_4|.$$

The table illustrates how cycle shape organises permutations by their underlying cycle structure; this will be essential when we define the signature homomorphism and study the alternating group $A_n$.

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Pre-Lecture Notes from University Notes

• Recall: for a homomorphism $\phi :G\to H$ with kernel $K=\ker \phi$, $K⊴G$ so the quotient $G/K$ is a well-defined group of cosets $gK$ and the canonical projection $\pi (g)=gK$ is a homomorphism.
• First isomorphism theorem: if $\phi :G\to H$ has kernel $K$ and image $I$, then $G/K\cong I$ via $\theta (gK)=\phi (g)$; proof checks that $\theta$ is well-defined, is a homomorphism, and is bijective.
• Example: the homomorphism $\phi :\mathbb{Z}\to {\mathbb{Z}}_n$ given by $\phi (m)=[m{]}_n$ has kernel $n\mathbb{Z}$ and image ${\mathbb{Z}}_n$, so $\mathbb{Z}/n\mathbb{Z}\cong {\mathbb{Z}}_n$ for all $n\in \mathbb{N}$.
• Second isomorphism theorem: for $H\le G$ and $N⊴G$ the map $\phi :H\to (HN)/N$, $\phi (h)=hN$, is a homomorphism with kernel $H\cap N$ and image $(HN)/N$, giving $H/(H\cap N)\cong (HN)/N$.
• Third isomorphism theorem (“Fool’s Cancellation”): for $N⊴G$ and $N\le M⊴G$, the map $\phi :G/N\to G/M$, $\phi (gN)=gM$, has kernel $M/N$ and image $G/M$, yielding $(G/N)/(M/N)\cong G/M$ and showing $M/N⊴G/N$.
• Definition: the cycle shape of $g\in {\mathrm{S}}_n$ is the ordered tuple of cycle lengths $(r_1,r_2,\dots ,r_m)$ when $g$ is written as a product of disjoint cycles with lengths $r_1\ge \cdots \ge r_m$, and the identity has cycle shape $\varnothing$.
• Worked examples: $g=(135729)(48)(1061211)\in {\mathrm{S}}_{12}$ has cycle shape $(6,4,2)$; $h=(12)(34)(78)(910)\in {\mathrm{S}}_{10}$ has cycle shape $(2,2,2,2)$; in ${\mathrm{S}}_4$ the possible cycle shapes are $\varnothing$, $(2)$, $(2,2)$, $(3)$, $(4)$ and account for all $4!=24$ permutations.
• Next time: define the signature homomorphism $\mathrm{sgn}:{\mathrm{S}}_n\to \{\pm 1\}$, use cycle shape to compute $\mathrm{sgn}(g)$ efficiently, and study the alternating group $A_n=\ker (\mathrm{sgn})$.