MTH3003 Weekly Problems 6

Original Documents: Problem Sheet / My Handwritten Solutions / mth3003 weekly problem sheet 6 solutions.pdf

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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 $x{y}^{-1}\in \mathrm{Im}(\theta )$.

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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.

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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.

Hint. Recall that multiplication in the quotient group $(HN)/N$ is given by $(h_{1}N)(h_{2}N)=h_1{h}_{2}N$. Now check the definition of a homomorphism.

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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

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

Hint. Recall from linear algebra that if $A,B,C$ are $n\times n$ matrices, then $\det (ABC)=\det (A)\det (B)\det (C)$, and for any invertible $A$ we have $\det (A^{-1})=\det (A{)}^{-1}$.

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