Understanding isomorphisms (abstract algebra) I have two unrelated questions on the topic of isomorphism.


*

*Why is it important to know that two groups are isomorphic? Are there other relations that we can draw from knowing that two groups are isomorphic aside from the fact that their structures are fundamentally the same?

*The lecture today was about finitely generated abelian groups, where the professor said that if $G$ were a finitely generated abelian group, it would be a direct product of cyclic groups (Fundamental theorem of finitely generated abelian groups)
Why is it a direct product here and not simply the product of cyclic groups?
 A: If two groups are isomorphic, you can essentially think of them as being THE SAME GROUP, so it really is quite a strong property.  They may be represented in different ways, for example as a collection of invertible matrices, or as a collection of permutations, but they have the same structure.
The direct product is usually what we mean when we talk about constructing groups from smaller pieces.  Describing as a product of subgroups is not especially clear, because important pieces of information are missing (whether the subgroups intersect nontrivially, whether they commute, etc.).  In this case it is what is known as an internal direct product, which is just a way to say the product of subgroups that all commute, and pairwise share only the identity of the group.
A: There is a great practical advantage of knowing when two groups (or other algebraic structures) are isomorphic. If you have a problem that is difficult to solve in one group but easy to solve in another, isomorphic group, you can transform the problem, solve it, and transform the solution back.
An example is the computation of the geometric mean of $n$ positive-real random numbers $(x_1,x_2,\ldots,x_n)$, defined as the $n$-th square root of the product of all $n$ numbers, i.e.
$$
\mathrm{GM} = \sqrt[n]{\prod\limits_{k=1}^{n} x_k}.
$$
Usually, multiplication is more difficult than addition, and taking the $n$-th square root is even harder. Fortunately, the multiplicative group of positive real numbers $(\mathbb{R}_{>0},\cdot)$ is isomorphic to the additive group of real numbers $(\mathbb{R},+)$, with $\ln : \mathbb{R}_{>0} \to \mathbb{R}$ being the isomorphism from $\mathbb{R}_{>0}$ to $\mathbb{R}$ and $\exp : \mathbb{R} \to \mathbb{R}_{>0}$ being the isomorphism from $\mathbb{R}$ to $\mathbb{R}_{>0}$. Noticing that $x = \exp(\ln(x))$ for all $x \in \mathbb{R}$, we can do the following:
$$
\sqrt[n]{\prod\limits_{k=1}^{n} x_k} = \underbrace{\exp\left( \ln\left( \sqrt[n]{\prod\limits_{k=1}^{n} x_k} \right) \right)}_\text{use isomorphisms} = \exp\left( \dfrac{1}{n}\ln\left( \prod\limits_{k=1}^{n} x_k \right) \right) = \exp\left( \dfrac{1}{n}\sum\limits_{k=1}^{n} \ln(x_k) \right)
$$
Note that $\dfrac{1}{n}\sum\limits_{k=1}^{n} \ln(x_k)$ is the arithmetic mean of the log-transformed numbers. To summarize, we observed that $(\mathbb{R}_{>0},\cdot)$ and $(\mathbb{R},+)$ are isomorphic, and could therefore simplify the computation of the geometric mean. It is nothing but the exp-transformed arithmetic mean of the log-transformed numbers.
A: Basically, as Math is considered, isomorphic groups ( or other structures) are equal to the effect of the situation at hand, and share all properties in the area in which they are isomorphic. One may say, that to the effects of a property, if$G,G'$ are isomorphic, that $G(G')$ is a relabeling of $G',(G)$ element-wise.
A: *

*Knowing isomorphisms can be very useful and can give you different ways of viewing a group which may be more enlightening. For example the symmetric group $S_3$ is isomorphic to the symmetries of the triangle $D_6$ but also to the group of matrices $GL_2(\mathbb{Z}/2)$. This also helps when trying to determine what group you have when you know certain properties.

*There are other types of products, for example semidirect products or the product group $HK$ (for subgroups $H$, $K \leq G$ which is a group if $HK=KH$), but in general you will lose the fact that the group is abelian. Taking direct product ensures the product is abelian by making the two subgroups commute with each other.
A: I like to think of a statement such as "the group $G$ is isomorphic to the group $H$" as being a statement about the measurement of symmetry, in the same way that "car $A$ has as many people in it as car $B$" is a statement about the measurement of number. If I have two baskets, one containing five oranges, the other five apples then, although I would never say it, I could regard the baskets as being "isomorphic" as $\textit{sets}$, because they contain the same $\textit{number}$ of items (even though the items are of different types).
In group theory, one of the examples I really like is the fact that $S_4$ is isomorphic to the group of rotations of a cube, under composition (consider the permutation action of $S_4$ on the four diagonals of the cube). In some sense, this says that the rotational symmetry of the cube is the $\textit{same}$ as, for example, the symmetry of rearrangements of four people in a queue, because these phenomena have the same $\textit{multiplicative}$ structure. And here, it is truly useful to know the isomorphism because, assuming you know that the transpositions generate $S_4$, then by the isomorphism you will know that 180-degree rotations about the edges generate all rotations of the cube.
As for the question on products of groups, there are a number of different ways to build a new group $G$ from existing groups $H$ and $K$. Usually, these constructions are such that $H$ and $K$ will be isomorphic to subgroups or quotient groups of the newly constructed $G$. Direct product is one relatively easy such construction, but there are others. Semi-direct product is a generalization of direct product; $\textit{wreath}$ product is a far-reaching generalization of semi-direct product. As has already been mentioned, there is also the notion of the subgroup product $HK$, where for this to make sense $H$ and $K$ must already be subgroups in a larger group. So to say simply "product" of cyclic groups is somewhat ambiguous (though such a statement will probably be clear in most contexts), but to know that cyclic groups are the building blocks of all finitely-generated abelian groups by $\textit{direct}$ product alone is very strong.
