What are the differences between rings, groups, and fields? Rings, groups, and fields all feel similar. What are the differences between them, both in definition and in how they are used?
 A: Any group $G$ is isomorphic to its opposite group $G^{\text{op}}$ via the map $g \mapsto g^{-1}$, however there is no such natural map for rings and in general it is not true that a ring is isomorphic to its opposite ring. 
Therefore, it is always possible to obtain a right action of a group $G$ if a left action is given whereas it may not be possible to equip a left $R$-module with a right $R$-module structure.
A: This answer aims to be a clear exposition of what groups, rings, and fields are, and how they are different. It does not, however, do much to motivate why we are interested in these algebraic structures in the first place. For that, I will refer the reader to Dummit and Foote's Abstract Algebra, which has a nice section at the beginning of part 1 giving an overview of the historical context surrounding groups. The proofs of the elementary properties of groups, rings, and fields mentioned in this post can also be found in Dummit and Foote, or indeed in any introductory abstract algebra textbook, or on Wikipedia.
Let $G$ be a set. A binary operation on $G$ is a function from $G\times G$ to $G$. Typically, we use infix notation to denote binary operations rather than prefix notation; that is, if the name of our binary operation is $\star$, then we write $a\star b$ instead of $\star(a,b)$. Intuitively speaking, a binary operation is a "rule" $\star$ that associates to every ordered pair $(a,b)$ of elements of $G$ an element $a\star b$ of $G$. For example, addition on $\mathbb N$ can be regarded as a binary operation that associates to each pair of natural numbers $(a,b)$ their sum $a+b$.
A group is an ordered pair $(G,\star)$, where $\star$ is a binary operation on $G$ satisfying the following "group axioms":

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*Associativity: for all $a,b,c\in G$,$$(a\star b)\star c=a\star(b\star c) \, .$$

*Existence of an identity element: there is a $b\in G$ such that for all $a\in G$, $$a\star b=b\star a=a \, .$$It can be proven from this axiom alone that there is only one $b\in G$ with the aforementioned property. It is called the identity element (or neutral element) of the group, and is denoted as $e$.

*Existence of inverse elements: for all $x\in G$, there is a $y\in G$ such that $$x\star y=y\star x=e \, .$$It can be proven using all three of the group axioms that for each $x$, this element $y$ is unique. It is called the inverse of $x$, and is denoted as $x^{-1}$.

If $(G,\star)$ is a group, then it is common to say that $G$ is a group under $\star$. In fact, if it is clear from context what the binary operation is, then it is common to speak even more loosely and call the set $G$ a group. When one speaks of the "elements of a group", this always refers to the elements of the set $G$.
Familiar examples of groups include $(\mathbb Z,+)$ and $(\mathbb Q\setminus\{0\},\times)$, where $+$ and $\times$ refer to the usual addition and multiplication operations, respectively. It is worth noting that in $(\mathbb Z,+)$, we follow a different notational convention to the one used above: the identity is written as $0$, and the inverse of $x$ is written as $-x$. This is known as additive notation, and it is common to use this whenever the group operation is written as $+$. In $(\mathbb Q\setminus\{0\},\times)$, we use multiplicative notation, where the identity is written as $1$, and the inverse of $x$ is sometimes written as $1/x$ instead of $x^{-1}$. Again, it is standard to use multiplicative notation whenenever the group operation is written using a symbol like $\cdot$ or $\times$.
Some non-examples of groups include $(\mathbb N,+)$ and $(\mathbb Q,\times)$. If we adopt the convention that $0$ is not a natural number, then it is easy to see why $(\mathbb N,+)$ is not a group: it doesn't contain an identity element. Even if we consider $0$ to be a natural number, $(\mathbb N,+)$ still fails to be a group: although $0$ acts as the identity element, every element of $\mathbb N$ other than $0$ does not have an inverse in $\mathbb N$. Note that $(\mathbb Q,\times)$ is "almost" a group: multiplication is associative, $1$ acts as the identity element, and every non-zero element has an inverse. However, $0$ itself does not have an inverse. Excluding $0$ gives us the group $(\mathbb Q\setminus\{0\},\times)$ mentioned earlier.
Here is a less obvious example of a group. Let $a$ and $b$ be any two distinct mathematical objects, and consider the ordered pair $(\mathcal P(\{a,b\}),\triangle)$, where $\triangle$ denotes the symmetric difference of two sets. I will leave it to the reader to verify that this is indeed an example of a group. This example also serves to reinforce the idea that the elements of a group need not be numbers. In fact, they can be any kind of mathematical object, including sets, matrices, functions, vectors, and so on.
All of the examples of groups shown thus far are not just examples of groups, but of abelian groups. An abelian group is a group with a commutative operation: for all $a,b\in G$,$$a\star b = b\star a \, .$$
In contrast to a group, a field has two operations. To be precise, a field is an ordered triple $(F,+,\times)$, where $+$ and $\times$ are binary operations on $F$ satisfying the following "field axioms":

*

*The ordered pair $(F,+)$ is an abelian group. This group is written additively; the element $0$ is called the additive identity of the field.

*The ordered pair $(F\setminus\{0\},\times)$ is an abelian group. This group is written multiplicatively; the element $1$ is called the multiplicative identity of the field.

*The operation $\times$ is distributive over $+$: for all $a,b,c\in F$,$$a\times(b+c)=(a\times b)+(a\times c)\quad\text{and}\quad(a+b)\times c=(a\times c)+(b\times c) \, .$$
The operation $+$ is called addition and $\times$ is called multiplication, even in cases when they are not the usual arithmetic operations that we are familiar with.
A ring is an ordered triple $(R,+,\times)$, where $+$ and $\times$ satisfy a list of "ring axioms". These axioms are identical to those of a field, except that we impose fewer requirements on the ordered pair $(R\setminus\{0\},\times)$: it now only has to be an associative structure, rather than an abelian group.
Every field is a ring, and the concept of a ring can be thought of as a generalisation of the concept of a field. Alternatively, a field can be conceptualised as a particular kind of ring, one whose non-zero elements form an abelian group under multiplication.
A ring with a multiplicative identity (i.e. an element $1$ such that $x\times 1 = 1\times x = x$ for all $x\in R$) is called a ring with identity or unital ring. As with many mathematical concepts, there are slight variations in how they are defined. Some authors require that all rings have an identity, and describe what we called a ring a rng (the omission of the letter i signals that it is unnecessary for there to be an identity).
A: You're right to think that the definitions are very similar. The main difference between groups and rings is that rings have two binary operations (usually called addition and multiplication) instead of just one binary operation. 
If you forget about multiplication, then a ring becomes a group with respect to addition (the identity is 0 and inverses are negatives). This group is always commutative!
If you forget about addition, then a ring does not become a group with respect to multiplication. The binary operation of multiplication is associative and it does have an identity 1, but some elements like 0 do not have inverses. (This structure is called a monoid.)
A commutative ring is a field when all nonzero elements have multiplicative inverses. In this case, if you forget about addition and remove 0, the remaining elements do form a group under multiplication. This group is again commutative.
A division ring is a (not necessarily commutative) ring in which all nonzero elements have multiplicative inverses. Again, if you forget about addition and remove 0, the remaining elements do form a group under multiplication. This group is not necessarily commutative. An example of a division ring which is not a field are the quaternions.
A: A group is an abstraction of addition and subtraction—except that the group operation might not be commutative.  But the important part is that there is an operation, which is something like addition, and the operation can be reversed, so there is also something like subtraction.
To this, a ring adds multiplication, but not necessarily division.
To this, a field adds division.
(Going the other way from a group, we have the monoid, which has addition, but not subtraction.)
A: I won't explain what a ring or a group is, because that's already been done, but I'll add something else.  One reason groups and rings feel similar is that they are both "algebraic structures" in the sense of universal algebra. So for instance, the operation of quotienting via a normal subgroup (for a group) and a two-sided ideal (for a ring) are basically instances of quotienting via an invariant equivalence relation in universal algebra.  A field, by contrast, is not really a construction of universal algebra (because the operation $x \to x^{-1}$ is not everywhere defined) -- which is why free fields don't exist, for instance -- though they are a special case of rings.  
A: They should feel similar! In fact, every ring is a group, and every field is a ring. A ring is an abelian group with an additional operation, where the second operation is associative and the distributive property make the two operations "compatible".
A field is a ring such that the second operation also satisfies all the properties of an abelian group (after throwing out the additive identity), i.e. it has multiplicative inverses, multiplicative identity, and is commutative.
