On isomorphism theorems.

This post follows last article, in this articles we discuss some properties about isomorphisms theorems, which are tightly connected to the definition of subgroups and quotient groups.

First, the first isomorphism theorem tells us that for any homomorphism $𝜑:𝐺\to {𝐺}^{\prime }$, we can decompose it into

Consider the part $𝐺\twoheadrightarrow 𝐺/ker𝜑$. This tells us a particular interesting thing, that every group with any homomorphism out of it carries a quotient group structure, especially when $𝜑$ is not injective and it’s also not trivial, which means that there are elements that got mapped to the same element in ${𝐺}^{\prime }$, and furthermore that $𝜑$ does not send all $𝐺$ to ${𝑒}_{{𝐺}^{\prime }}$, we shall have a non trivial quotient group

$$𝐺/ker𝜑$$

because if $𝜑\left(𝑎\right)=𝜑\left(𝑏\right)={𝑔}^{\prime }$, then we may naturally consider the set ${𝜑}^{-1}\left({𝑔}^{\prime -1}\right)$, then we have $𝑎\left[{𝜑}^{-1}\left({𝑔}^{\prime -1}\right)\right]\subseteq ker𝜑$ and $𝑏\left[{𝜑}^{-1}\left({𝑔}^{\prime -1}\right)\right]\subseteq ker𝜑$, since $𝑎\ne 𝑏$, this creates non trivial kernel and therefore non trivial quotient group.

The reason why $𝐺/ker𝜑\stackrel{\sim }{⟶}im𝜑$ is an isomorphism also contains an interesting intuition. We know that a map is not an isomorphism when it’s not injective or not surjective, but by definition of $im𝜑$, this map is clearly surjective. Furthermore, we can consider the quotient group $𝐺/ker𝜑$ have “collapsed” all elements that can possible cause any non-injectivity: if $𝑎\ne 𝑏\in 𝐺$ and $𝜑\left(𝑎\right)=𝜑\left(𝑏\right)\in {𝐺}^{\prime }$, then this information will be recorded by $ker𝜑$ and therefore quotient out $ker𝜑$ removes the non-injectivity. Actually, it’s fair easy to see that any group homomorphism $𝜑$ determines an equivalence relation that is compatible with the group structure. And $ker𝜑$ is the subgroup that records the “null displacement” w.r.t. this equivalence relation.

Now we come to a closely-related lemma of the third isomorphism theorem

We don’t care about the inclusion part, what I want to justify is the actuall important thing of this lemma. The lemma itself looks nonsensical upon first read. But actually it shows that Any subgroup of $𝐺/𝐻$ is also a quotient group $𝐾/𝐻$ where $𝐻\le 𝐾\le 𝐺$. Notice that when giving the inverse, we define

$$𝐾={𝑢}^{-1}\left({𝐾}^{\prime }\right)=\left\{𝑎\in 𝐺|𝑎𝐻\in {𝐾}^{\prime }\right\}$$

we now show that in fact, ${𝐾}^{\prime }=𝐾/𝐻$, notice that $𝐾/𝐻$ is defined to be

$$\begin{array}{cc}& 𝐾/𝐻\\ =& \left\{𝑘𝐻|𝑘\in 𝐾\right\}\\ =& \left\{𝑘𝐻|𝑘\in \left\{𝑎\in 𝐺|𝑎𝐻\in {𝐾}^{\prime }\right\}\right\}\\ =& {𝐾}^{\prime }\end{array}$$

as $𝑘\in 𝐺$ is just the elements in $𝐺$ that makes $𝑘𝐻\in {𝐾}^{\prime }$, the definition $\left\{𝑘𝐻|𝑘\in \left\{𝑎\in 𝐺|𝑎𝐻\in {𝐾}^{\prime }\right\}\right\}$ automatically collapse to just ${𝐾}^{\prime }$. Since ${𝐾}^{\prime }$ is just any subgroup of $𝐺/𝐻$ we pick, we can conclude that any subgroup of $𝐺/𝐻$ is itself a quotient group. And this quotient group has a one-to-one correspondence with subgroups $𝐾$ of $𝐺$ containing $𝐻$, thus we may have the following diagram (notice this is not a commutative diagram in the tranditional sense, as arrows are not functions)

That is, each subgroup $𝐾$ of $𝐺$ containing $𝐻$ uniquely determines a subgroup $𝐾/𝐻$ of $𝐺/𝐻$ and vice versa in a surjective sense that there are “no other subgroups of $𝐺$ and $𝐺/𝐻$ not described by this correspondence”.