Isomorphism Theorems

Groups

Theorem A

Suppose $f$ is a group homomorphism $f:G \to H$, then

1. $\ker(f)$ is a normal subgroup of $G$
2. ${\rm im}(f)$ is a subgroup of $H$
3. ${\rm im}(f)$ is isomorphic to $G/\ker(f)$.

Proof

1. Let $u,v \in \ker(f)$. Then $$f(uv^{-1}) = f(u)f(v^{-1}) = f(u)f(v)^{-1} = 1 \cdot 1 = 1.$$ So $\ker(f)$ is a subgroup. Suppose, for contradiction, that $\ker(f)$ is not normal. That is, let $a \in \ker(f)$ and $b \in G$ such that $b^{-1} a b \not\in \ker(f)$. Then $$f(b^{-1} a b) = f(b^{-1}) f(a) f(b) = f(b^{-1}) f(b) = 1,$$ which implies $b^{-1} a b \in \ker(f)$. This contradicts our hypothesis.
2. Let $u, v \in {\rm im}(f)$. By definition of image, there exist some $a,b \in G$ such that $f(a)=u,$ $f(b)=v.$ Then $$f(ab^{-1})=f(a)f(b^{-1})=f(a)f(b)^{-1}=uv^{-1}$$ and $f(ab^{-1})$ is clearly in the image of $f$.
3. Let $\sim$ be the relation defined as $$a \sim b \iff f(a) = f(b)$$ (1) $\sim$ is an equivalence relation: this is inherited from equality. (2) $\sim$ is a congruence: suppose $a \sim b$ and $v \sim u$. Then $$f(av) = f(a)f(v) = f(b)f(u) = f(bu),$$ so $av \sim bu$ as needed. Partition $G$ by $\sim$ (this constructs $G/\ker(f)$ explicitly) and note that any two elements in the same partition are mapped by $f$ to the same element of ${\rm im}(f)$, so the map $\varphi:G/\ker(f) \to {\rm im}(f)$, sending $[g]$ to $f(g)$, is well defined. Now, let $[p],[q] \in G/\ker(f)$ and note that since $\sim$ is a congruence, $[p][q]$ is well-defined and is equal to $[pq]$. Moreover $$\varphi([pq]) = f(pq) = f(p)f(q) = \varphi([p])\varphi([q]),$$ so $\varphi$ is a homomorphism. $\varphi$ is injective by construction, since if $\varphi([a])=\varphi([b])$, then $f(a)=f(b)$ and hence $[a]=[b]$. It is surjective, since any $a \in {\rm im}(f)$ has a preimage in $G$ which belongs to one of the equivalence classes. So $a=\varphi([a])$.
   $\blacksquare$

Intuition

A homomorphism gives rise to a congruence on . Since it is a congruence, the quotient is well-defined. By construction, sending any representative of some equivalence class in through covers all of with no overlaps (we collapsed these using ) and no missed spots (by definition of ). This gives an isomorphism.

Theorem B

Let $S, N$ be subgroups of a group $G$ with $N$ normal. Then

1. The product $SN$ is a subgroup of $G$.
2. $N$ is a normal subgroup of $SN$
3. The intersection $S \cap N$ is a normal subgroup of $S$
4. $SN/N \cong S/(S \cap N)$

Proof

1. Let $s_1, s_2 \in S$, $g_1, g_2 \in N$ and let $g = g_1g_2^{-1} \in N$ and $s = s_1s_2^{-1} \in S$. Then $$s_1 g_1 (s_2 g_2)^{-1} = s_1g_1g_2^{-1}s_2^{-1} = s_1gs_2^{-1},$$ but since $N$ is normal, there exists $g' \in N$ such that $gs_2^{-1} = s_2^{-1}g'$ and hence $$s_1gs_2^{-1} = s_1s_2^{-1}g' = sg'.$$ so $SN$ passes the subgroup test.
2. Indeed, conjugating an element of $N$ by any element of $G$ produces an element of $N$, so in particular conjugating an element of $N$ by an element of $SN$ produces an element of $N.$
3. Suppose $g \in S$ and $g \in N$. Let $a \in S$ and consider $a^{-1}ga$. Clearly this is an element of $S$, but also note that since $N$ is a normal subgroup of $G$, it has to be an element of $N$ as well. Thus, conjugating any element of $S \cap N$ by an element of $S$ produces an element of $S \cap N$, so it is a normal subgroup.
4. Let $Z := S \cap N$. Let $s \in S, n \in N$ and consider the coset $snN \in SN/N$. Clearly, $snN = sN$. Now, if $s \in N$ then $sN = N$, the converse is also true, since $sN = N$ in particular implies $s \cdot 1 = s \in N$ . Now, take the map $f:S \to SN/N$ sending $s$ to $sN.$ Clearly $f$ is surjective, by the earlier argument. Note that $f$ is a group homomorphism, since $$f(s_1s_2) = (s_1s_2)N \overset{!}{=} s_1N \cdot s_2N = f(s_1)f(s_2),$$ where the marked equality holds because $N$ is normal. By the first isomorphism theorem, that gives us an isomorphism ${\rm im}(f) = SN/N \cong S/\ker(f)$ and this kernel is precisely $Z$.
   $\blacksquare$

Intuition

The idea is that given $SN$, quotienting out $N$ kills all of $N$ and this includes the overlap with $S$. This can be seen clearly from cosets: if $n \in N$ and $s \in S$, $snN = sN$ and if $s$ also happens to be in $N$, then $sN = N$.

Modules

Theorem B

Proposition

Let $N$, $P$ be submodules of an $R$-module $M$. Then

• $N + P$ is a submodule of $M;$
• $N \cap P$ is a submodule of $P$ and $$\frac{N+P}{N} \cong \frac{P}{N\cap P}$$

Proof

First, observe that $N + P$ is a subgroup of $M$. Any element of $N+P$ is already an element of $M$, and since $R$-action on $N + P$ is inherited from $M$, it is sufficient to make $N + P$ into a module, provided that closure holds.

Let $n \in N$ and $p \in P$. For an arbitrary $r \in R$, consider $r(n + p) = rn + rp$. Since $N$ is a submodule, $rn \in N$. Similarly, $rp \in P$. This demonstrates that $rn + rp \in N + P$, as required.

Similarly, for $N \cap P$ we only need to check closure. Let $u \in N \cap P$ and take an arbitrary $r \in R$. Since $u \in N$, $ru \in N$ as well. Similarly, $ru \in P$. So $N \cap P$ is a submodule.

Let $K = N \cap P$ and consider the map $\varphi : (N + P) \to P/K$ defined as $n + p \mapsto p + K.$ It is clearly a surjective homomorphism. So ${\rm im}(\varphi) = P/K$ and $\ker(\varphi) = N$.

Let’s check that $\varphi$ is well-defined. Let $n_1 + p_1 = n_2 + p_2$. Note that $n_1 - n_2 = p_1 - p_2$ and since $n_1 - n_2 \in N$ and $p_1 - p_2 \in P$, we know that this difference lives in $K$.

Finally, the canonical decomposition

gives the required isomorphism.

$\blacksquare$

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