## Properties of homomorphisms of abelian groups

Let $\phi: B\to A$ be a homomorphism of abelian groups $(B,+)$ and $(A,+)$ (we denoted operations in both groups by the same symbol – these are different operations, but no confusion will arise; you will always see from the context in which group we work; same for 0s in these groups).

The image of $\phi$ is the set $Im(\phi)=\{\phi(b)\mid b\in B\};$
the kernel of $\phi$ is the set $Ker(\phi)=\{b\in B\mid \phi(b)=0\}.$

It is easy to check (check this!) using the definition of homomorphism that sets just defined are subgroups: $Im(\phi)\leq A$ and $Ker(\phi)\leq B.$

Note that image and kernel are concepts parallel to the concepts of the image and kernel of a linear map between vectorspaces; in particular such linear maps provide homomorphisms between additive groups of the corresponding vectorspaces.

Cosets and quotient, a.k.a. factor, groups .

Let $H\leq B.$ Given $b\in B,$ the set $b+H:=\{b+h\mid h\in H\}$ is called the coset of $b$ in $B.$
For any $x,y\in B$ the following holds: either $x+H=y+H,$ or $x+H\cap y+H=\emptyset.$ (Indeed, if $x-y\in H,$ then we have the former, as $H$ is closed under addition, otherwise the latter – check this.) Thus we have a partition of $B$ into cosets of $H.$

A fancy way of saying the latter is to say that being in the same coset of $H$ defines an equivalence relation on $B.$
As an example, consider $B=\mathbb{Z}$ and a positive integer $m.$ Then the cosets of $H=m\mathbb{Z}$ in $B$ are the sets of integers of the form $n+mk,$ with $k,n\in\mathbb{Z}.$ Then $n_1+mk_1$ and $n_2+mk_2$ are in the same coset of $H$ iff $n_1-n_2+m(k_1-k_2)\in H,$ i.e. when $m$ divides $n_1-n_2.$ Recalling the procedure of division with reminder, is easy to see that the cosets are $x+H,$ where $x=0,1,\dots, m-1.$

The set of cosets of $H$ in $B$ has a natural structure of group defined on it.

The quotient group $B/H=(\{b+H\mid b\in B\},+_{B/H})$ has the set of cosets of $H$ in $B$ as the set of elements; the addition is defined by $(x+H)+_{B/H}(y+H):=(x+y)+H.$

In a way, we can identify the operation $+_{B/H}$ with the $+$ in $B.$
The former is the same as the latter taken “modulo $H$“.
So we still need to check that this way one ideed has an abelian group. Commutativity and associativity in $B/H$ is immediate from the associativity and commutativity in $B.$ Then, $0_{B/H}=H$ (note that $H=0+H$ is a coset of $H$ in $B$, just any other coset), and $-(x+H)$ is $-x+H.$
Continuing the above example $B=\mathbb{Z}, H=m\mathbb{Z}$ we have $B/H$ isomorphic to $\mathbb{Z}_m.$ This is in fact not suprising, and is implied by the following general theorem.

1st Isomorphism Theorem.

Let $\phi: B\to A$ be a homorphism of abelian groups. Then we can identify $Im(\phi)$ with certain quotient group of $B.$ Namely $Im(\phi)\cong B/Ker(\phi).$ Specifically, the latter isomorphism is given by $b+Ker(\phi)\mapsto\phi(b).$

The proof of this is left as an easy exercise (that you should do!). The (rather unfortunate) name (1st…) comes from the fact that there are more similar results in group theory (sometimes also the already mentioned facts that $Ker$ and $Im$ are subgroups is included as its part).
It follows that every $H\leq B$ is the kernel of the homomorphism $B\to B/H.$ What distinguishes such homomorphisms from the general ones is that they are surjective, i.e. the target of the mapping equals its image.
(Thus there is a 1-1 correspondence between the surjective homomorphisms from $B$ and subgroups of $B$.)

A question to think about: what are the groups $B/H$ in the cases when $H=B,$ and $H={0}?$

Armed with the 1st isomorpshism theorem, we can easily give a proof of our claim about cyclic groups. Namely, let $A=\langle s\rangle$ be a cyclic group. Consider a homomorphism $\phi:\mathbb{Z}\to A$ defined by $\phi(n)=ns.$
Trivially, $\phi$ is surjective. Let $m$ the minimal positive integer such that $ms=0,$ provided that such an integer exists. We claim that if $m$ does not exist then $\phi$ is an isomorphism, and otherwise $Ker(\phi)=\langle m\rangle.$ Of course, if $ks=0$ for a negative $k$, then $-ks=0,$ too, so there is no loss in generality in assuming $m\ge 0,$ if it exists. Thus in the case of no such $m$ we indeed have an isomorphism. In the case when such an $m$ exists it is obvious that $m\mathbb{Z}=Ker(\phi)$ (check this!). Thus, by the 1st isomorphism theorem, $A\cong \mathbb{Z}/m\mathbb{Z},$ completing the proof of the claim.

Sometimes, although not always, a subgroup isomorphic to $B/H$ is readily found inside $B.$ This is for instance always the case when $B$ is the additive group of a finite-dimensional vectorspace, and $H$ the additive group of a subspace. Then $B=A\oplus H,$ as is known from linear algebra.
(A fancy way to say the latter is to say that $B$ is a split extension of $H.$)
You are encouraged to show that this is indeed not true in general (Hint: consider $B=\mathbb{Z}_4$ and $\mathbb{Z}_2\cong H=2B\le B.$) (Such extensions are called non-split.)

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### 6 Responses to “Properties of homomorphisms of abelian groups”

a typo on the paragraph before 1st isomorphism theorem. it should be B/H instead of H/B.

Shouldn’t A be isomorphic to Z/mZ instead of Z/Z_m in the second last paragraph?

• Dima Says:

and for this one, too. Indeed, $latex \mathbb{Z}_m$ isn’t even a subgroup of $\mathbb{Z}$

3. Liu Mei Says:

Under Cosets and Quotients, given b in B, the set b+H is called the coset of H, not of b, right?

• Dima Says:

b+H is a coset of H (not the coset, as there could be more than one!) with a representative b.
In view of this, b+H is the coset of b.