Confusion about equality: mathematical objects versus the symbols that describe them Something about the notion of equality bothers me. Let's consider an example from group theory, the group with two elements $\mathbb{Z}/2\mathbb{Z}=\{0,1\}$ under the binary operation satisfying $0+0=0$ and $0+1=1+0=1$.
Based on my understanding of equality, the statement $(1+1)+1=1$, contains no mathematical content beyond $1=1$, since the group element $(1+1)+1$ literally is the group element $1$. This bothers me...
It seems to me that there are two mathematical structures in question. The group $G=\mathbb{Z}/2\mathbb{Z}$ which is the set $\{0,1\}$ coupled with a binary function (subset of $\big((G\times G),G\big)$) which is the set $\{\big((0,0),0\big),\big((0,1),1\big),\,\big((1,0),1\big)\}$. This group structure seems to induce (or maybe the other way around?) a new mathematical structure describing the behavior of the symbols used to describe $G$. Perhaps this new mathematical structure $\mathcal{G}$, could be thought of as equivalence classes of words in $\mathbb{Z}/2\mathbb{Z}$ under the natural operation. Then, the statement $[(1+1)+1]=[1]$ where $[1]\in\mathcal{G}$ and $[(1+1)+1]\in\mathcal{G}$ are both equivalence classes of words in $G$ actually has real mathematical content, namely that $(1+1)+1$ and $1$ are in the same equivalence class.
When I read the statement $(1+1)+1=1$, I want to be able to mathematically (not just mentally) interpret equality as $(1+1)+1$ and $1$ are distinct representations of the same group element.
Is this point of view wrong/unhelpful? Is there some field that studies questions like this? If so, I would love textbook recommendations (early graduate level). Based on my very limited knowledge, I wonder if this is related to model theory or the theory of formal languages. Feel free to retag.
 A: 
the statement $(1+1)+1=1$, contains no mathematical content beyond $1=1$, since the group element $(1+1)+1$ literally is the group element $1$. This bothers me...

It should; it's wrong. This idea was exploded by Frege in the 19th century. People keep saying it anyway (including on this site) but it's clearly wrong.   Consider:

My uncle Jerry is the Vice-President of Advertising for CBS.

Uncle Jerry and the VP are, as you put it, literally the same individual, so on the traditional account, this statement is devoid of content.  But it clearly isn't devoid of content; it tells you that my uncle Jerry is the Vice-President of Advertising for CBS, something you didn't previously know.
Or to take a mathematical example:
$$e^{ix} = \cos x + i\sin x$$
Is this "devoid of mathematical content"?  The two expressions have "literally the same value".  Why is that Euler guy famous, when all his famous theorems are vacuous?
In Frege's account, there are two kinds of meaning, which he calls "sense" and "reference". The two expressions $1$ and $(1+1)+1$ have the same value, which Frege calls “reference”: they both refer to the element $1$ of $\Bbb Z/2\Bbb Z$.  But the expressions have different "senses"; they're saying different things.
In more modern language, we might say: there's a difference between a computation and the value that it computes.  There are many ways to compute the same value, many paths to the same destination.  A great many "interesting" statements of mathematics can be understood as observations that there are two different ways to compute the same value.
Mathematical expressions denote computations, not values. This is why we work so hard to evaluate them: we want the values!
A: I know there are probably many different perspectives to this, but I'll try to lay out the one I'm most familiar with here.
In formal first-order logic, there's always a clear distiction made between syntax and semantics. Terms are syntactic objects; they can be built out of variables like "$x$", constant symbols like "$1$" and function/operation symbols like "$+$", but have on their own no meaning beyond the syntax trees they represent. For example, "$(1+1)+1$" and "$1$" are two terms, and they have clearly different syntax trees, so we say they're syntactically unequal.
However, we haven't yet said anything about what exactly the symbols "$1$" and "$+$" we used mean. To actually give that term semantic meaning, we need to choose something for those symbols to refer to; we need to choose a structure to interpret that in. That structure $(A,f,a)$ can be any set $A$ with a binary operation $A\times A\to A$ to interpret "$+$" as and an element $a\in A$ to interpret "$1$" as; $(\mathbb Z/2\mathbb Z,+_{\mathbb Z/2\mathbb Z},[1])$ would be one example, but we could also choose $(\mathbb Z,+_{\mathbb Z},1)$, $(\mathbb N,+_\mathbb N,1)$ or any other. Since the terms "$(1+1)+1$" and "$1$" both evaluate to $[1]$ in the first case we say that they are semantically equal in $(\mathbb Z/2\mathbb Z,+,[1])$, but not for example in $(\mathbb Z,+,1)$. Similarly, if we also introduce a constant symbol "$2$", "$1+1$" and "$2$" are syntactially different but semantically equal in $(\mathbb Z,+,1,2)$.
Another notion of equality maybe worth mentioning comes from the fact that, of course, mathematicians don't always want to add in a new constant symbol for something they can already uniquely define; instead, they'll just define the term "$2$" to mean the term "$1+1$". "$1+1$" and "$2$" would then for example be not just semantically but also definitionally equal in $(\mathbb Z,+,1)$, while terms like "$2+1$" and "$1+2$" are equal only semantically but not by definition.
A: Consider the function $x \mapsto x+1$, ie the "add 1 function", in the group that you have fixed. That $(x \mapsto x+1)(1)$ = 0 is a result of the typical beta reduction.
More generally given a formal language, we might wish to say that two expressions, although not syntactically equal, are semantically equal ("1+1" and "2" being the prime candidates). We do so by defining a notion of primitive equality on the terms in general. This is often introduced as beta-reduction- the syntactic expression of a function applied to its input is semantically equal to the syntactic expression obtained via computation thereof.
For more, please see a book on the lambda calculus.
