This result is quite important through the rest of the exercises.

- We want to prove that $\varphi(x^n) = \varphi(x)^n$ for $n \in \mathbb{Z}^{+}$. By induction on $n$:
- Base case $n = 1$: $\varphi(x) = \varphi(x)$
- Ind case. Let true for $n$. Now $$ \varphi(x^{n + 1}) = \varphi(x^{n} x) = \varphi(x)^n \varphi(x) = \varphi(x)^{n + 1} $$

- Next for $n = -1$. This is a bit tricky. I can't find $\varphi(1_G) = 1_H$ stated anywhere in the book prior to this exercise. Since this result is missing, this is solved in a somewhat strange way. First $$ \varphi(x^n x^{-1}) = \varphi(x)^n \varphi(x)^{-1} $$ for $n \geq 2$. This we have proved in (a). Secondly $$ \varphi(x^n x^{-1}) = \varphi(x)^n \varphi(x^{-1}) $$ Now $$ \varphi(x)^{n} \varphi(x^{-1}) = \varphi(x)^{n} \varphi(x)^{-1} \Rightarrow \varphi(x^{-1}) = \varphi(x)^{-1} $$ by the cancellation law. Here we can make $n$ concrete. That is choose $n = 3$ or something like that. Finally we can deduce that since $$ \varphi((x^n)^{-1}) = \varphi(x^n)^{-1} = (\varphi(x)^{n})^{-1} $$ we have that $\varphi(x^n) = \varphi(x)^n$ for all $n \in \mathbb{Z} - 0$.

One important special case of the above is that $$ \varphi(1_G) = \varphi(x x^{-1}) = \varphi(x) \varphi(x)^{-1} = 1_H $$ from this follows that $\varphi(x^n) = \varphi(x)^n$ for all $n \in \mathbb{Z}$. An alternative proof is as follows: $$ 1_H = \varphi(1_G) \varphi(1_G)^{-1} = \varphi(1_G) \varphi(1_G) \varphi(1_G)^{-1} = \varphi(1_G) $$

Let $\varphi : G \rightarrow H$ be an isomorphism. We want to prove that $|\varphi(x)| = |x|$ for $x \in G$. We have from (2) and since $\varphi$ is well defined, that
$$
x^n = 1_G \Rightarrow \varphi(x^n) = \varphi(1_G) \Rightarrow \varphi(x)^n = 1_H
$$
We have from (2) and since $\varphi$ is a bijection, that
$$
\varphi(x)^n = 1_H \Rightarrow \varphi(x^n) = \varphi(1_G) \Rightarrow x^n = 1_G
$$
Here we use that $\varphi$ is a bijection and therefore injective in the right most implication.

Now due to what we have proved, and due to that an isomorphism implies a one to one correspondence between $\varphi(x)$ and $x$, we have that $G$ and $H$ has the same number of elements of some order $n$.

If $\varphi$ only a homomorphism, it is either not surjective or injective. If not surjective, we could have some $h \in H$ with no corresponding $x \in G$. If not injective, different elements of different order might be mapped to one element of one order. That is $x_0 \mapsto \varphi(x_0)$ and $x_1 \mapsto \varphi(x_0)$, where $|x_0| \neq |x_1|$, but $|\varphi(x_1)| = |\varphi(x_1)|$.

Let $\varphi : G \rightarrow H$ isomorphism. Now we want to prove that $G$ abelian iff $H$ abelian.

- Right direction. Let $G$ abelian. We have $$ ab = ba \Rightarrow \varphi(ab) = \varphi(ba) \Rightarrow \varphi(a) \varphi(b) = \varphi(b) \varphi(a) $$ since $\varphi$ is well defined.
- Left direction. Let $H$ abelian. We have $$ \varphi(a) \varphi(b) = \varphi(b) \varphi(a) \Rightarrow \varphi(ab) = \varphi(ba) \Rightarrow ab = ba $$ since $\varphi$ is injective. Here we don't care of $\varphi$ being surjective. That is $G$ might be a proper subgroup of $H$.

As stated above, we don't care whether $\varphi$ is surjective or not.

In the multiplicative group $\mathbb{R} - 0$ all elements except $-1$ has infinite order. $-1$ has order $2$. Now in $\mathbb{C}$ we have that $i^2 = -1 \Rightarrow i^4 = 1$. Thus $i$ has order $4$ which no element of $\mathbb{R} - 0$ has. Hence not isomorphic as of (c) on page 38.

Since $\mathbb{Q}$ is countable, wheras $\mathbb{R}$ is not, we have that $|\mathbb{R}| \gt |\mathbb{Q}|$. Hence these two groups are not isomorphic as of (a) on page 38.

This is a bit philosophical. But we have $$ \frac{0}{1},\frac{0}{2},\frac{0}{3},... $$ in $\mathbb{Q}$, all of which has order $1$. In $\mathbb{Z}$ we only have the element $1$ of this order. Hence by exercise (2), we have that these two groups are not isomorphic. The only problem is that I'm not sure whether that infinite series of fractions counts as the same element.

In $Q_8$ $i,j,k$ has order $4$. In $D_8$ only $r,r^3$ has this order. Hence $Q_8$ are not isomorphic to $D_8$.

We have that $$ n \neq m \Rightarrow n! \neq m! \Rightarrow |S_n| \neq |S_m| $$ Hence $S_n$ is not isomorphic to $S_m$.

We need to prove 3 things:

- This follows since $\theta$ is a bijection. That is $f_1 = \sigma \circ \theta^{-1}$ is a well defined function from $\Omega$ to $\Delta$ (since bijection). And $f_2 = \theta \circ \sigma$ is a well defined function from $\Delta$ to $\Omega$ (since $\theta$ itself is well defined).
- $\varphi$ is a bijection. Pick $$ \varphi^{-1}(\delta) = \theta^{-1} \circ \delta \circ \theta $$ For $\delta \in S_{\Omega}$. Now $$ \varphi^{-1}(\varphi(\sigma)) = \varphi^{-1}(\theta \circ \sigma \circ \theta^{-1}) = \theta \theta^{-1} \sigma \theta^{-1} \theta = \sigma $$ It goes the same for $\varphi(\varphi^{-1}(\sigma))$ - specifically since $\theta$ is a bijection. Hence two-sided.
- $\varphi$ is a homomorphism. We have $$ \varphi(\sigma \circ \tau) = \theta \circ \sigma \circ \tau \circ \theta^{-1} = \theta \circ \sigma \circ \theta^{-1} \circ \theta \circ \tau \circ \theta^{-1} = \varphi(\sigma) \circ \varphi(\tau) $$

Let $A$ and $B$ be groups. Define $\varphi : A \times B \rightarrow B \times A$ as $$ \varphi((a,b)) = (b,a) $$ For $\varphi$ we have $$ \varphi((a_1,b_1)(a_2,b_2)) = \varphi((a_1 \star a_2,b_1 \diamond b_2)) = (b_1 \diamond b_2,a_1 \star a_2) $$ we also have that $$ \varphi((a_1,b_1)) \varphi((a_2,b_2)) = (b_1,a_1)(b_2,a_2) = (b_1 \diamond b_2,a_1 \star a_2) $$ so $\varphi$ is clearly a homomorphism. It's also self inverse, ie $$ \varphi(\varphi((a,b))) = (a,b) $$ Hence $\varphi$ is an isomorphism, and we have that $$ A \times B \cong B \times A $$

Let $A,B,C$ groups. Let $G = A \times B$ and $H = B \times C$. We want to prove that direct products are associative, eg $G \times C \cong A \times H$. Pick $$ \varphi((a,(b,c))) = ((a,b),c) $$ for $a \in A,b \in B,c \in C$. So $\varphi$ is a homomorphism. We have: $$ \varphi((a_1,(b_1,c_1)) (a_2,(b_2,c_2))) = \varphi((a_1 a_2,(b_1 b_2,c_1 c_2))) = ((a_1 a_2,b_1 b_2),c_1 c_2) $$ Not to mention: $$ \varphi((a_1,(b_1,c_1))) \varphi((a_2,(b_2,c_2))) = ((a_1,b_1)(a_2,b_2),c_1 c_2) = ((a_1 a_2,b_1 b_2),c_1 c_2) $$ Pick $$ \varphi^{-1} (((a,b),c)) = (a,(b,c)) $$ Now $$ \varphi^{-1}(\varphi((a,(b,c)))) = (a,(b,c)) $$ and $$ \varphi(\varphi^{-1}(((a,b),c))) = ((a,b),c) $$ Hence $\varphi^{-1}$ is a two-sided inverse, and $\varphi$ is bijective, and $$ G \times C \cong A \times H $$ as desired.

Let $G,h$ groups. Let $\varphi : G \rightarrow H$ homomorphism. We want to prove that for the image we have $\varphi(G) \leq H$, that is is a subgroup. We do it as done in exercise 26 of section 1. That is by showing groups axioms, but restricted to $H_0 = \varphi(G)$.

- Associativity: sure
- Identity element: We have that $\varphi(1_G) = 1_H$, so this element exists in $H_0$.
- Inverse element: We have that $\varphi(x^{-1}) = \varphi(x)^{-1}$ for every $x \in G$ (note that since $G$ is a group, we have that $x \in G \Rightarrow x^{-1} \in G$). So $H_0$ is closed under taking inverses.

Now let $\varphi$ injective. Since subgroup, it should be reasonable to assume that $|H_0| \leq |G|$. But since injective (and well defined) we must have $|H_0| = |G|$. That is $\varphi$ is a bijection, and we have that $$ H_0 \cong G $$

Let $G,H$ groups. Let $\varphi : G \rightarrow H$ homomorphism. Define kernel, $K$, of $\varphi$ as: $$ \{ g \in G | \varphi(g) = 1_H \} $$ we want to prove that the kernel is a subgroup of $G$. That is $K \leq G$. As above:

- Associativity: sure
- Identity: Since we have at least that $\varphi(1_G) = 1_H$, we can just pick $1_G$ to be identity in $K$.
- Inverses: We have that $$ \varphi(x) = 1_H = \varphi(x) \varphi(x)^{-1} = \varphi(x) \varphi(x^{-1}) $$ which implies that $$ 1_H = \varphi(x^{-1}) $$ So for every $x \in K$ we have a $x^{-1} \in K$.

Now we want to prove that $\varphi$ injective iff $K = \{1_G\}$

- Left direction: Let $\varphi$ be injective. Since we have that $\varphi(1_G) = 1_H$ and that $1_H$ is unique, no other element can map through $\varphi$ than $1_G$. Thus $K = \{1_G\}$.
- Right direction: Let $K = \{ 1_G \}$. Here we can just spell out the map: $1_G \mapsto 1_H$. Quite boring, but it sure is injective.

Define $\pi : \mathbb{R}^2 \rightarrow \mathbb{R}$ by $\pi((x,y)) = x$. This exercise I have a hard time grasping. The structure isn't given, so assume none. Since no structure, sure $\pi$ is structure preserving. But we can't find the kernel since the existence of an identity element sort of implies that we have some operator to use it on. Another way to interpret is this: Let $H = (\mathbb{R},\star)$ be a group and let $G = H \times H$ be the direct product of this group with it self - that is a group. Let $\pi : G \rightarrow H$ be given as above. Now $$ \pi((x_1,y_1)(x_2,y_2)) = \pi(x_1 x_2 , y_1 y_2) = x_1 x_2 $$ and $$ \pi((x_1,y_1)) \pi((x_2,y_2)) = x_1 x_2 $$ So $\pi$ is a homomorphism. The kernel is $(1_H,y)$ for $y \in \mathbb{R}$.

Let $A,B$ groups. Let $G = A \times B$ - their direct product. Prove that the following maps are homomorphism and find their kernels

- $\pi_1 : G \rightarrow A$ given as $\pi_1((a,b)) = a$. We have $$ \pi_1((a_1,b_1)(a_2,b_2)) = \pi_1((a_1 a_2,b_1 b_2)) = a_1 a_2 $$ and $$ pi_1((a_1,b_1)) \pi_1((a_2,b_2)) = a_1 a_2 $$ So homomorphism. The kernel is as above $(1_A,b)$ for $b \in B$. Note that $\pi_1((a,b)) \neq 1_A$ when $a \neq 1_A$.
- $\pi_2 : G \rightarrow B$ given as $\pi_2((a,b)) = b$. We have $$ \pi_2((a_1,b_1)(a_2,b_2)) = \pi_2((a_1 a_2,b_1 b_2)) = b_1 b_2 $$ and $$ \pi_2((a_1,b_1)) \pi_2(a_2,b_2) = b_1 b_2 $$ thus a homomorphism. The kernel is $(a,1_B)$.

*This $\pi$ function is usually called a projection. So in functional programming we normally have $fst$ and $snd$ for $\pi_1$ respectively $\pi_2$.

Let $G$ group. Let the map $f : G \rightarrow G$ be given by $f(g) = g^{-1}$. We want to prove that this map is a homomorphism iff $G$ abelian. We do so

- Left direction: Let $f$ homomorphism. Now $$ f(ab) = (ab)^{-1} = b^{-1}a^{-1} \land f(ab) = f(a)f(b) = a^{-1}b^{-1} \Rightarrow b^{-1}a^{-1} = a^{-1}b^{-1} $$ An alternative proof using exercise 1.1.18: $$ (ab)^{-1} (ab) = 1 \Rightarrow f(ab) ab = 1 \Rightarrow f(a)f(b) ab = 1 \Rightarrow a^{-1}b^{-1} ab = 1 $$ or using same exercise: $$ a = ba(ba)^{-1} a = ba f(ba) a = ba f(b) f(a) a = ba b^{-1} a^{-1} a = bab^{-1} $$
- Right direction: Let $G$ abelian. Now $$ f(ab) = (ab)^{-1} = b^{-1}a^{-1} = a^{-1}b^{-1} = f(a)f(b) $$

Since inverses have inverses, $a$ and $a^{-1}$ ought to range over the same number and kind of elements.

Almost same setting as above. Let $G$ group. Let $f : G \rightarrow G$ be given as $f(g) = g^2$. We want to prove that $f$ homomorphism iff $G$ abelian:

- Left direcion. Let $f$ homomorphism. Now $$ f(ab) = abab \land f(ab) = f(a)f(b) = aabb \Rightarrow ba = ab $$ by right and left cancellation laws.
- Right direction. Let $G$ abelian. Now $$ f(ab) = abab = aabb = f(a)f(b) $$

Let $G = \{ z \in \mathbb{C} | \exists n \in \mathbb{Z}^{+} : z^n = 1\}$. We want to prove that for any fixed integer $k \gt 1$, the map $f : G \rightarrow G$ given as $f(z) = z^k$ is a surjective homomorphism, but not injective. Let's start by proving that $f$ is a homomorphism. Note the following: $G$ commutes since $(\mathbb{C},\star)$ does. I'm somewhat sure that any group that is based on the set $\mathbb{C}$, commutes. So let's go: $$ f(z_1 z_2) = (z_1 z_2)^{k} = z_1^k z_2^k = f(z_1) f(z_2) $$ Now I'm not at all sure why $f$ is surjective. Say we have some element $i$ of order $4$. Fix $k = 4$. Now $f(i^m) = 1$ for any $m$, thus the element $i$ doesn't seem to be the target of $f$ in any way. Thus not surjective. At least this is an argument for $f$ not being injective. Might need to get back to this one.

Let $G$ group. Let $Aut(G)$ as described in the assignment. Now we go right ahead:

- Associative: Not now
- Identity: Pick $id \in Aut(G)$. $id$ is a isomorphism - a quite boring one, though. Now $$ id \circ f = f \circ id = f $$ for $f \in Aug(G)$.
- Inverse: A isomorphism is a bijection. That is it has a two sided inverse, $f^{-1}$. So pick this: $$ f \circ f^{-1} = f^{-1} \circ f = id $$

$id : G \rightarrow G$ an isomorphism? Let's go. Homomorphism: $$ id(a b) = ab = id(a) id(b) $$ Now $id$ is a two-sided self inverse. That is $$ id(id(a)) = a $$ Done!

Let $f : \mathbb{Q} \rightarrow \mathbb{Q}$ be given by $f(q) = kq$ for $k \neq 0, k \in \mathbb{Q}$. We want to prove that $f$ is an automorphism (an isomorphism from a group to it self). We start by proving $f$ to be a homomorphism: $$ f(q_1 q_2) = k q_1 q_2 $$

Let $A$ abelian group. Let $f(a) = a^k$ for some fixed $k \in \mathbb{Z}$. We want to prove that $f$ is a homomorphism (isomorphism for $k = -1$):

- $k \gt 0 \lor k \lt 0$: We have that $$ f(ab) = (ab)^k = a^k b^k = f(a) f(b) $$ where we exploit that $A$ abelian during the second equality.
- $k = 0$: We have that $$ f(ab) = (ab)^0 = 1 = 1 1 = a^0 b^0 = f(a) f(b) $$
- $k = -1$. We have that $$ f(ab) = (ab)^{-1} = b^{-1} a^{-1} = a^{-1} b^{-1} = f(a) f(b) $$ so in order to show that that $f$ be homomorphic, we don't need the abelian part here. Now here $f$ is a two-sided self inverse: $$ f(f(a)) = a $$ so $f$ is isomorphic.
- Note that for $k = 1$ we have $f$ as an automorphism as well.

Let $G$ finite group. Let $\sigma$ automorphism. Let $\sigma(g) = g \iff g = 1$. Let $\sigma^2$ be the identity map from $G$ to $G$. We want to prove that $G$ abelian. As of the hint we first want to show that any element in $G$ can be written on the form $x^{-1} \sigma(x)$. There seems two ways (not very formal compared to the rest of the book) to show this. One is by constructing a function $f : G \rightarrow G$ by $$ f(y) = x \in G : x^{-1} \sigma(x) = y $$ where $$ f(1) = 1 $$ and show that $f$ is a bijection. First of all we need to restrict $f$ in order to make it well defined. That is if we for $x_1 \neq x_2$ have for some $y$ $$ x_1^{-1} \sigma(x_1) = y = x_2^{-1} \sigma(x_2) $$ we need to choose the one of the $x$'s that isn't used yet. Since $x$ and $y$ are from the same set, we can map each $y$ uniquely to a $x$ and vice versa. Since we have a correspondence between the $y$'s and the $x$'s, and since $\sigma$ is a bijection, we have that $f$ is surjective. Furthermore we have that $$ x_1^{-1} \sigma(x_1) = x_2^{-1} \sigma(x_2) \Rightarrow y_1 = y_2 $$ and thus $f$ is injective.

Another way seems to amount to show that $$ x_1 = x_2 \iff x_1^{-1} \sigma(x_1) = x_2^{-1} \sigma(x_2) $$ But with some added details. I'm not going further into this.

Back to the proof. We have each element of $G$ on the form $x^{-1} \sigma(x)$. By applying $\sigma$ we get $$ \sigma(x^{-1} \sigma(x)) = \sigma(x^{-1}) \sigma^2(x) = ... $$

Let $G$ finite group. Let $x,y \in G$ be distinct of order 2 that generetes $G$. We want to prove that $G$ is isomorphic to $D_{2n}$, with $n = |xy|$. Since $D_{2n}$ is finite, by exercise 1.2.6 we have that $x,t = xy$ satisfy the same relations in $G$ that $r,s$ do in $D_{2n}$. If $x,y$ generates $G$, then since $ty = x$, so does $t,y$. Now construct $\varphi$ as follows $$ \varphi(y) = s, \varphi(t) = r, \varphi(1) = 1 $$ And all other elements of $G$ is deconstructed by $\varphi$ into powers of $t$ and $y$ and send into $D_{2n}$ in the form of powers of $r$ respectively $s$, where the given element is reconstructed. The deconstruction and reconstruction conforms to the relations of $G$ and $D_{2n}$. Since some element is uniquely constructed by generators of a group, we have that $\varphi$ inherently is an isomorphism. That is for $\varphi^{-1}$ that deconstructs in $D_{2n}$ and reconstructs in $G$ (mechanical inverse to $\varphi$), we have $$ \varphi(g) = d \iff \varphi^{-1}(d) = g $$ for $g \in G$ and $d \in D_{2n}$. Now we have that $$ \varphi(g_1 g_2) = d_1 d_2 = \varphi(g_1) \varphi(g_2) $$ since the product $g_1 g_2$ is uniquely deconstructed. And we have that $$ G \cong D_{2n} $$

I haven't had anything to do with linear algebra for quite some time. This exercise is postponed.

Same as above.

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