MTH3003 Weekly Problems 6

Original Documents: mth3003 weekly problem sheet 6.pdf / mth3003 weekly problem sheet 6 handwritten solutions.pdf / mth3003 weekly problem sheet 6 solutions.pdf

Vibes: Four short proofs around the homomorphism machinery: image is a subgroup, signature gives a $C_2$ quotient, restriction-to-quotient maps are homomorphisms, and a determinant case study using all three Isomorphism theorems. Mostly Quick-Subgroup-Test or First-Isomorphism-Theorem applications.

Used Techniques:

• Quick Subgroup Test (identity, closure, inverses).
• First Isomorphism Theorem: surjective $\theta :G\to H\Rightarrow G/\ker \theta \cong H$.
• Third Isomorphism Theorem: $G/N/H/N\cong G/H$ when $N⊴H⊴G$.
• Determinant as a homomorphism $\det :{\mathrm{GL}}_2(\mathbb{R})\to {\mathbb{R}}^{\ast }$.

---

6.1. Image of a Homomorphism is a Subgroup

Question

Let $G$ and $H$ be groups, and let $\theta :G\to H$ be a group homomorphism. Prove that $\mathrm{Im}(\theta )$ is a subgroup of $H$ (that is, $\mathrm{Im}(\theta )\le H$).

Hint. Use the Quick Subgroup Test: show that $\mathrm{Im}(\theta )$ is nonempty and that for all $x,y\in \mathrm{Im}(\theta )$ we have $x{y}^{-1}\in \mathrm{Im}(\theta )$.

Solution. Apply the Quick Subgroup Test directly. Fix $h_1,h_2\in \mathrm{Im}(\theta )$, so $h_i=\theta (g_i)$ for some $g_i\in G$.

• Identity. $\theta (e_G)=e_H$, so $e_H\in \mathrm{Im}(\theta )$.
• Closure. $\theta (g_1{g}_2)=\theta (g_1)\theta (g_2)=h_1{h}_2$, so $h_1{h}_2\in \mathrm{Im}(\theta )$.
• Inverses. $\theta (g_1^{-1})=\theta (g_1{)}^{-1}=h_1^{-1}$, so $h_1^{-1}\in \mathrm{Im}(\theta )$.

Hence $\mathrm{Im}(\theta )\le H$. $■$

---

6.2. Using the Signature to Get a Quotient Isomorphic to $C_2$

Question

Suppose $n\ge 2$. Show that there exists a surjective homomorphism $\sigma :S_n\to C_2$. Prove that $S_n$ has a normal subgroup $K$ such that $S_n/K\cong C_2$.

Hint. Use the signature function $\sigma$ as the homomorphism in the First Isomorphism Theorem.

Solution. The Signature function $\sigma :S_n\to \{\pm 1\}\cong C_2$ is a surjective group homomorphism (any transposition has signature $-1$, identity has $+1$). Let $K=\ker (\sigma )=A_n$, the Alternating group of even permutations. By the First Isomorphism Theorem,

$$\boxed{S_n/A_n\cong \mathrm{Im}(\sigma )=C_2.}$$

So $K=A_n$ is the required normal subgroup. $■$

---

6.3. Natural Map from a Subgroup into a Quotient

Question

Let $G$ be a group with $H\le G$ and $N⊴G$. Define a map $\phi :H\to (HN)/N$ by $\phi (h)=hN$ for each $h\in H$. Prove that $\phi$ is a group homomorphism.

Solution. Fix $h_1,h_2\in H$. Compute

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

using the multiplication rule in the quotient group. Hence $\phi$ is a homomorphism. $■$

---

6.4. Normal Subgroups in ${\mathrm{GL}}_2(\mathbb{R})$ via Determinant

Question

Consider the group ${\mathrm{GL}}_2(\mathbb{R})$ of all invertible $2\times 2$ real matrices, the subgroup ${\mathrm{SL}}_2(\mathbb{R})$ of invertible $2\times 2$ real matrices with determinant equal to $1$, and the subgroup

$$P=\{M\in {\mathrm{GL}}_2(\mathbb{R}):\det (M)>0\}.$$

1. Prove that $P$ and ${\mathrm{SL}}_2(\mathbb{R})$ are normal subgroups of ${\mathrm{GL}}_2(\mathbb{R})$.
2. Give an intuitive description of the quotient groups ${\mathrm{GL}}_2(\mathbb{R})/{\mathrm{SL}}_2(\mathbb{R})$, $P/{\mathrm{SL}}_2(\mathbb{R})$, and ${\mathrm{GL}}_2(\mathbb{R})/P$.
3. Prove that

$${\mathrm{GL}}_2(\mathbb{R})/{\mathrm{SL}}_2(\mathbb{R})/P/{\mathrm{SL}}_2(\mathbb{R})\cong {\mathrm{GL}}_2(\mathbb{R})/P.$$

Write $G={\mathrm{GL}}_2(\mathbb{R})$, $K={\mathrm{SL}}_2(\mathbb{R})$.

Part 1: $P,K⊴G$. For any $M,A\in G$,

$$\det (M^{-1}AM)=\det (M{)}^{-1}\det (A)\det (M)=\det (A),$$

since determinants are real numbers and multiplication is commutative in ${\mathbb{R}}^{\ast }$. Hence:

• If $A\in P$, $\det (M^{-1}AM)=\det (A)>0$, so $M^{-1}AM\in P$. So $P⊴G$.
• If $A\in K$, $\det (M^{-1}AM)=\det (A)=1$, so $M^{-1}AM\in K$. So $K⊴G$.

$\boxed{P⊴G,K⊴G.}$

Part 2: Quotient descriptions. The determinant map $\det :G\to {\mathbb{R}}^{\ast }$ (the multiplicative group of nonzero reals) is a surjective homomorphism with $\ker (\det )=K$. By the First Isomorphism Theorem,

$$G/K\cong {\mathbb{R}}^{\ast }\text{(non-zero reals under multiplication)}.$$

Restricting $\det$ to $P$ gives $\det :P\to {\mathbb{R}}_{>0}$, surjective with kernel $K$, so

$$P/K\cong {\mathbb{R}}_{>0}\text{(positive reals under multiplication)}.$$

For $G/P$: matrices split by sign of determinant. Take $J=\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)$, $\det (J)=-1$. Any $M\in G$ has $\det (M)>0$ ($M\in P$) or $\det (M)<0$ (then $JM\in P$). The cosets are $\{P,JP\}$ with $J^2=I$, giving

$$G/P\cong \{I,J\}\cong C_2.$$

Part 3: Apply the Third Isomorphism Theorem. Since $K⊴P⊴G$ (with $K⊴G$ already shown):

$$\frac{G/K}{P/K}\cong G/P.■$$

The chain of quotients telescopes - concretely ${\mathbb{R}}^{\ast }/{\mathbb{R}}_{>0}\cong C_2$, recovering the sign of the determinant.