MTH3003 Weekly Problems 5

Original Documents: Problem Sheet / [[mth3003 weekly problem sheet 5 handwritten solutions.pdf|My Handwritten Solutions]] / Provided Solutions

Vibes: More conceptual than Weeks 3 and 4, but the arguments are short once you lean on the definitions and the standard homomorphism propositions (identity, powers, kernel).

Used Techniques:

• homomorphism definition: $\theta (g_1{g}_2)=\theta (g_1)\theta (g_2)$.
• Standard consequences: $\theta (e_G)=e_H$ and $\theta (g^m)=(\theta (g){)}^m$.
• Kernel method: show $g^{-1}kg\in \ker (\theta )$ for all $g\in G$, $k\in \ker (\theta )$.
• isomorphism definition: a bijective homomorphism.

---

5.1. Checking Homomorphisms and Isomorphisms for Squaring Maps

Question

Determine whether the following maps are homomorphisms or isomorphisms, providing brief justifications rather than full proofs:

1. $\phi :C_7\to C_7$ defined by $\phi (g)=g^2$ for all $g\in C_7$.
2. $\phi :S_7\to S_7$ defined by $\phi (g)=g^2$ for all $g\in S_7$.

1. $\phi :C_7\to C_7$, $\phi (g)=g^2$: this is a homomorphism because $C_7$ is cyclic (hence Abelian), so $(xy{)}^2=x^2{y}^2$ in $C_7$. It is an isomorphism because if $C_7=⟨g⟩$, then $\phi (g)=g^2$ is also a generator (since $2$ is invertible modulo $7$), so $\phi$ is bijective.

2. $\phi :S_7\to S_7$, $\phi (g)=g^2$: this is not a homomorphism in general because squaring does not respect products in a non-Abelian group. Counterexample in $S_7$: let $a=(12)$ and $b=(23)$, so $a^2=b^2=e$ but $(ab)=(123)$ and $(ab{)}^2=(132)\ne e$, hence $\phi (ab)\ne \phi (a)\phi (b)$. Since it is not a homomorphism, it cannot be an isomorphism (an isomorphism must be a homomorphism).

---

5.2. Identity Preservation under Homomorphisms

Question

Given groups $G$ and $H$ with a homomorphism $\theta :G\to H$, prove that $\theta (e_G)=e_H$.

Fix any $g\in G$.

Then $\theta (g)=\theta (e_{G}g)=\theta (e_G)\theta (g)$ by the definition of homomorphism.

Right-multiply by $\theta (g{)}^{-1}$ in $H$ to obtain $e_H=\theta (e_G)$.

---

5.3. Exponentiation Preservation under Homomorphisms

Question

Given groups $G$ and $H$ with a homomorphism $\theta :G\to H$, prove that $\theta (g^m)=(\theta (g){)}^m$ holds for all $g\in G$ and $m\in \mathbb{N}$.

We prove by induction on $m\in \mathbb{N}$.

Base case $m=1$: $\theta (g^1)=\theta (g)=(\theta (g){)}^1$.

Inductive step: assume $\theta (g^m)=(\theta (g){)}^m$; then

$$\theta (g^{m+1})=\theta (g^{m}g)=\theta (g^m)\theta (g)=(\theta (g){)}^m\theta (g)=(\theta (g){)}^{m+1},$$

Using the homomorphism property and the inductive hypothesis.

(As noted on the sheet and in the lecture notes, this extends to all integers $m$ as well.)

---

5.4. Normality of the Kernel

Question

Given groups $G$ and $H$ with a homomorphism $\theta :G\to H$, prove that $\ker (\theta )⊴G$. You may assume that $\ker (\theta )$ is a subgroup of $G$.

Assume $\ker (\theta )\le G$.

To show $\ker (\theta )⊴G$, fix $g\in G$ and $k\in \ker (\theta )$, and compute

$$\theta (g^{-1}kg)=\theta (g{)}^{-1}\theta (k)\theta (g).$$

This uses the homomorphism property together with the standard fact that homomorphisms preserve inverses (which follows from the “powers” result).

Since $k\in \ker (\theta )$, we have $\theta (k)=e_H$, so $\theta (g^{-1}kg)=\theta (g{)}^{-1}{e}_H\theta (g)=e_H$, hence $g^{-1}kg\in \ker (\theta )$.

Therefore, $\ker (\theta )$ is normal in $G$.

Optional: $\ker (\theta )\le G$

Identity: $\theta (e_G)=e_H$, so $e_G\in \ker (\theta )$. Closure: if $a,b\in \ker (\theta )$ then $\theta (ab)=\theta (a)\theta (b)=e_H{e}_H=e_H$. Inverses: if $a\in \ker (\theta )$ then $\theta (a^{-1})=\theta (a{)}^{-1}=e_H^{-1}=e_H$.

---

5.5. Isomorphism between Dihedral and Symmetric Groups

Question

Prove that $D_6$ is isomorphic to $S_3$ (that is, $D_6\cong S_3$.

Recall the presentations on the sheet:

$$D_6=\{e,\rho ,{\rho }^2,\sigma ,\sigma \rho ,\sigma {\rho }^2\},{\rho }^3=e,{\sigma }^2=e,\rho \sigma =\sigma {\rho }^{-1},$$

and

$$S_3=\{e,(123),(132),(12),(13),(23)\}.$$

Define $\theta :D_6\to S_3$ by specifying images of generators:

$$\theta (\rho )=(123),\theta (\sigma )=(12),$$

And then extending using $\theta ({\rho }^2)=\theta (\rho {)}^2$ and $\theta (\sigma {\rho }^i)=\theta (\sigma )\theta (\rho {)}^i$.

Check the defining relations are preserved:

• $\theta (\rho {)}^3=(123{)}^3=e$, matching ${\rho }^3=e$.
• $\theta (\sigma {)}^2=(12{)}^2=e$, matching ${\sigma }^2=e$.
• $\theta (\rho )\theta (\sigma )=(123)(12)=(13)$ and $\theta (\sigma )\theta (\rho {)}^{-1}=(12)(132)=(13)$, so $\theta (\rho \sigma )=\theta (\sigma {\rho }^{-1})$.

So $\theta$ is a homomorphism from $D_6$ to $S_3$ (it respects the relations that define products of $\rho$ and $\sigma$).

Now compute the images of all six elements:

$$\theta (e)=e,\theta (\rho )=(123),\theta ({\rho }^2)=(132),\theta (\sigma )=(12),\theta (\sigma \rho )=(13),\theta (\sigma {\rho }^2)=(23).$$

These are exactly all the elements of $S_3$, so $\theta$ is surjective.

Since $|D_6|=6$ and $|S_3|=6$