Exploration 4: Homomorphisms

• Finite order means there’s some integer $n$ such that $g^n=1$.
• Then $\sigma(g)^n=\sigma(g^n)=\sigma(1)=1$. Since $\sigma(g)^n=1$, the order of $\sigma(g)$ must divide $n$.

• We prove the contrapositive.
• Let $m$ be the order of $g\in G$ and $n$ be the order of $\sigma(g)\in H$. If $\sigma$ doesn’t preserve the order of $g$, then $n<m$ with $m$ being a multiple of $n$, from the last theorem.
• Since $n$ is the order of $\sigma(g)$ and $m$ is a multiple of $n$, we have $\sigma(g)^m=\sigma(g)^n=e$, implying that both $g^m=e$ and $g^n\ne e$ map to the identity element $e$.
• Therefore $\sigma$ is not injective.

• The proof is by induction on the positive formula $\phi$. It can be one of three things:
  • Equalities $a=b$, possibly composed together with $\land$ and $\lor$ (but not $\lnot$)
  • An existential: $\phi=\exists x\ldotp \psi(x)$ for some positive formula $\psi(x)$
  • A universal: $\phi=\forall x\ldotp \psi(x)$ for some positive formula $\psi(x)$
• The base case is when the positive formula is an equality $a=b$. The terms on both sides only feature group product and inverses, which are both preserved by all homomorphisms, so equalities are preserved.
• The first inductive case is when the positive formula is an existential: $\phi=\exists x\ldotp \psi(x)$ for some positive formula $\psi(x)$. If $\phi$ is true then $\psi(g)$ is true for some element $g\in G$. By induction, $\sigma$ preserves $\psi(x)$, so $\psi(\sigma(g))$ is true in $H$ as well. Then we know that there is some element $\sigma(g)\in H$ that satisfies $\phi=\exists x\ldotp \psi(x)$, so $\phi$ is preserved as well.
• The second inductive case is when the positive formula is a universal: $\phi=\forall x\ldotp \psi(x)$ for some positive formula $\psi(x)$. If $\phi$ is true then $\psi(g)$ is true for all elements $g\in G$. By induction, $\sigma$ preserves $\psi(x)$, so $\psi(\sigma(g))$ is true in $H$ as well. Since $\sigma$ is surjective, we know that $\sigma(g)$ can be every element in $H$, so every $\sigma(g)\in H$ satisfies $\phi=\forall x\ldotp \psi(x)$, so $\phi$ is preserved as well.

Since automorphisms are injective, they only map elements to elements of the same order. In the case of unique order elements, those elements must map to themselves.

If $g$ is fixed by all inner automorphisms $\sigma_a$, it means $aga^{-1}=g$, or $ag=ga$, for all $a$. In other words, $g$ commutes with all elements $a$ and therefore is in the center.

For $g_1,g_2\in G$ to differ by the kernel of the homomorphism, we must have $g_1g_2^{-1}\in\ker\sigma$, which is equivalent to saying $\sigma(g_1g_2^{-1})=e_H$, i.e. $\sigma(g_1)=\sigma(g_2)$. Thus, two elements that differ by an element in $\ker\sigma$ are equal under $\sigma$.

By definition. Since surjective means the whole codomain is mapped to, it’s the same as saying that the mapped subset of $H$ is all of $H$.

• ($\to$) If $\sigma(a)=\sigma(b)$, then $\sigma(ab^{-1})=e_H$ implies $ab^{-1}=e_G$ since only $e_G$ maps to $e_H$. But then $a=b$, therefore $\sigma$ is injective.
• ($\from$) Since injective homomorphisms preserve the order of each element, and the only order $1$ element in any group is $e$, only $e_G$ gets mapped to $e_H$, so the kernel is trivial.

• ($\to$) If $\im\sigma\iso G$, then by the isomorphism every element of $\im\sigma$ is necessarily mapped to once, which means every element of $H$ is mapped to at most once, therefore $\sigma$ is injective.
• ($\from$) The elements mapped to by $\sigma$ are exactly $\im\sigma$. And since $\sigma$ is injective, they are mapped to exactly once, therefore $G\to\im\sigma$ is an isomorphism.

• Let $k$ denote elements of the kernel $\ker\sigma$.
• The kernel is a subgroup because:
  • $\sigma(k_1k_2)=\sigma(k_1)\sigma(k_2)=ee=e$ implies $k_1k_2$ is in the kernel, and
  • $\sigma(k^{-1})=\sigma(k)^{-1}=e^{-1}=e$ implies $k^{-1}$ is in the kernel.
• The kernel is invariant under conjugation:
  • $\sigma(gkg^{-1})=\sigma(g)\sigma(k)\sigma(g^{-1})=\sigma(g)e\sigma(g^{-1})=e$ implies $gkg^{-1}$ is in the kernel for arbitrary $g$ in the group.
• So the kernel is invariant under conjugation, therefore normal.

• The image is closed under product and inverse, therefore a subgroup of the codomain:
  • Product: $\sigma(g_1)\sigma(g_2)=\sigma(g_1g_2)$
  • Inverse: $\sigma(g)^{-1}=\sigma(g^{-1})$
    

• Since $\tilde{\phi}\circ\pi(g)=\phi(g)$ implies $\tilde{\phi}([g])=\phi(g)$, the homomorphism $\tilde{\phi}$ is forced to be the map $[g]\mapsto \phi(g)$, and therefore this map is unique.
• $\tilde{\phi}$ is well-defined: we need to show that given $[g]=[h]$, $\phi(g)=\phi(h)$.
  • From $[g]=[h]$ we know $[g^{-1}h]=[e]$, so $g^{-1}h$ is in $H$. Since the kernel contains $H$, $\phi(g^{-1}h)=e$.
  • Then $\phi(g)=\phi(g)\phi(g^{-1}h)=\phi(h)$.
• $\tilde{\phi}$ preserves product and inverse: they are basically inherited from $\phi$.
  • $\tilde{\phi}([a][b])=\tilde{\phi}([ab])=\phi(ab)=\phi(a)\phi(b)$
  • $\tilde{\phi}([a]^{-1})=\tilde{\phi}([a^{-1}])=\phi(a^{-1})=\phi(a)^{-1}$
• Therefore $\tilde{\phi}$ is a homomorphism.
• To show that it is injective, we use the same argument as the one used for well-definedness, but going in the opposite direction.
  • Assuming $\phi(g)=\phi(h)$, we know $e=\phi(g^{-1}h)$.
  • This implies $g^{-1}h$ is in the kernel of $\tilde{\phi}$, so $[g^{-1}h]=[e]$, which implies $[g]=[h]$.
• Therefore $\tilde{\phi}$ is an injective homomorphism such that $\tilde{\phi}\circ\pi=\phi$.

• First, we restrict the codomain of $\sigma$ to its image $\im\sigma$, obtaining a surjective homomorphism $\phi:G\to\im\sigma$.
• Apply the universal property of the quotient to $\phi$. Then there is an injective homomorphism $\tilde{\phi}:G/\ker\phi\to\im\phi$, such that $\tilde{\phi}\circ\pi=\phi$.
• By definition, $\pi$ and $\phi$ are both surjective. Because $\tilde{\phi}\circ\pi=\phi$, we know $\tilde{\phi}$ must be surjective as well.
• Since $\tilde{\phi}$ is both injective and surjective, it is an isomorphism between $G/\ker\phi$ and $\im\phi$.
• Since $\phi$ is the same as $\sigma$ but with a restricted codomain, they have the same kernel and image. Therefore $G/\ker\sigma\iso\im\sigma$.

Since the image of $\sigma$ is necessarily $\{e\}$ (homomorphisms only map identity to identity) we know that the kernel of $\tau$ is $\{e\}$ as well, and homomorphisms with trivial kernel are injective.

Since the kernel of $\tau$ is necessarily $\{B\}$ (because everything from $B$ got mapped to $e$) we know that the image of $\sigma$ is $B$ as well, implying $\sigma$ is surjective.

• Since $\sigma$ is injective, $A\iso\im\sigma$, By exactness, we have $\im\sigma=\ker\tau$, therefore $A\iso\ker\tau$. The kernel of $\tau$ is a normal subgroup of its domain $B$.

• From the previous proof we know that in a short exact sequence, $\ker\tau\iso A$. If $\sigma$ is an inclusion map, then $\ker\tau=A$.
• Recall the first isomorphism theorem for $\tau$ is $\im\tau\iso B/\ker\tau$. In this case, we have
  • $\ker\tau=A$.
  • $\im\tau=C$ since the last map maps to $\{e\}$.
• Then $C\iso B/A$.

• Left inverse homomorphisms are homomorphisms $\sigma^{-1}$ such that $\sigma^{-1}\circ\sigma=\id_A$.
• Right inverse homomorphisms are homomorphisms $\tau^{-1}$ such that $\tau\circ\tau^{-1}=\id_C$.
• If we have $\sigma^{-1}:B\to A$, then:
  • Elements of $B$ are writable as the product of some element in $\ker\sigma^{-1}$ and some element in $\im\sigma$. Proof: $$\begin{aligned} &~\sigma^{-1}(b(\sigma\sigma^{-1}b)^{-1})\\ =&~\sigma^{-1}(b\sigma((\sigma^{-1}b)^{-1}))\\ =&~\sigma^{-1}b \sigma^{-1}\sigma((\sigma^{-1}b)^{-1})\\ =&~\sigma^{-1}b (\sigma^{-1}b)^{-1}\\ =&~e\\ \end{aligned}$$ so we have $b(\sigma\sigma^{-1}b)^{-1}\in\ker\sigma^{-1}$. Also, we have $\sigma\sigma^{-1}b\in\im\sigma$. The product of the two is $b$.
  • This means we can write elements of $B$ as a product of elements of $\ker\sigma^{-1}$ and $\im\sigma$, but it’s only unique if the two subgroups share no elements in $B$ (besides identity).
  • Elements $k\in\ker\sigma^{-1}$ satisfy $\sigma^{-1}(k)=e$. If $k$ is also in the image of $\sigma$, then $k=\sigma(a)$ for some $a\in A$, then $\sigma^{-1}(k)=e$ becomes $\sigma^{-1}(\sigma(a))=e$ which implies $a=e$ and therefore $k=\sigma(a)=e$. So the only element shared by $\ker\sigma^{-1}$ and $\im\sigma$ is $e$.
  • Therefore we can uniquely write elements of $B$ as a product of elements of $\ker\sigma^{-1}$ and $\im\sigma$, and thus $B$ is a direct product of the two.
  • By exactness, $\im\sigma$ is isomorphic to $A$. What’s left to prove is that $\ker\sigma^{-1}$ is isomorphic to $C$.
  • By exactness, $\im\sigma=\ker\tau$, and $\tau$ is surjective. So every $c\in C$ is equal to $\tau(ki)$ where $ki\in B$, $k\in\ker\sigma^{-1}$ and $i\in\im\sigma=\ker\tau$. Since $i\in\ker\tau$ this is equal to $\tau(k)$. Therefore every $c=\tau(k)$ for some $k\in\ker\sigma^{-1}$, so we have a bijection $\ker\sigma^{-1}\to C$, so they are isomorphic.
  • Since $B$ is isomorphic to a direct product of $A\times C$, the sequence is split.
• The argument for $\tau^{-1}:C\to B$ is essentially identical to the above.