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In my study of groups, rings, modules etc, I've seen the three isomorphism theorems stated and proved many times. I use the first one ( $G/\ker \phi \cong \operatorname{im} \phi$ ) very often, but I can't recall having ever used the other two. Can anyone give some examples where they are used in a crucial way in some proof?

For clarity, let us say that the 2nd one is : $(M/L)/(N/L) \cong M/N$ under the appropriate conditions, and the 3rd one is $(M+N)/N \cong M/(M\cap N).$

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In Galois theory, I think they could have plenty of occasions of applications. Notice that you could view the second isomorphism theorem as an associativity of homomorphisms, IIRC. – awllower Feb 20 '13 at 9:30
@Katie: You will come across thousands of applications when you continue studying algebra. – Martin Brandenburg Feb 20 '13 at 11:37
3rd theorem: counting dimensions of vector spaces. – wj32 Feb 20 '13 at 12:18
@awllower, care to elaborate on that? Aren't homomorphisms always associative, since they are arrows in the category of groups? I don't see what the second isomorphism theorem would do with that. – Sintrastes Jan 11 at 20:06
@Sintrastes Consider the surjective homomorphism $M\rightarrow M/L\rightarrow (M/L)/(N/L)$ whose kernel is just N. – awllower Jan 11 at 23:19
up vote 3 down vote accepted

As Boris and Alexander have mentioned, both these theorems are used a lot in the study of normal series, and particularly soluble groups.

One basic example of this is the following. A group $G$ is metabelian if it has a normal subgroup $N$ such that $N$ and $G/N$ are both abelian.

Theorem Any subgroup of a metabelian group is metabelian.

Proof Let $G$ be a metabelian group with normal abelian subgroup $N$ such that $G/N$ is abelian. Let $H$ be any subgroup of $G$. Then $H\cap N$ is a normal abelian subgroup of $H$ and $H/(H\cap N)\cong HN/N$, which is a subgroup of $G/N$ and is hence abelian. Thus $H$ is metabelian. $\Box$

The same sort of idea could be used with induction to prove that any subgroup of a soluble group of derived length $n$ is also soluble, of derived length at most $n$, although I don't think that's the standard proof.

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Thanks, seeing how it was used in that theorem really helped me appreciate the importance of that isomorphism theorem! – Katie Dobbs Feb 21 '13 at 12:06
Oh, great! I wasn't really sure this answer would add anything much to what's already been said here, but I thought I might as well put it up anyway. – Tara B Feb 21 '13 at 12:12

This is an application of the second isomorphism theorem, although the theorem does not play a crucial role in it.

Let $a, b$ be positive, say, integers. Then $$ a \mathbf{Z} + b \mathbf{Z} = \gcd(a, b) \mathbf{Z}, $$ and $$ a \mathbf{Z} \cap b \mathbf{Z} = \operatorname{lcm}(a, b) \mathbf{Z}. $$ Now the second isomorphism theorem gives you the isomorphism $$ \frac{\gcd(a, b) \mathbf{Z}}{b \mathbf{Z}} = \frac{a \mathbf{Z} + b \mathbf{Z}}{b \mathbf{Z}} \cong \frac{a \mathbf{Z}}{a \mathbf{Z} \cap b \mathbf{Z}} = \frac{a \mathbf{Z}}{\operatorname{lcm}(a, b) \mathbf{Z}}. $$ Comparing orders you get $$ \frac{b}{\gcd(a, b)} = \frac{\operatorname{lcm}(a, b)}{a}, $$ which is the well-known formula $$ \gcd(a, b) \operatorname{lcm}(a, b) = a b. \tag{gcd/lcm} $$ Clearly (gcd/lcm) can be proved without recourse to the second isomorphism theorem. But whenever I teach the theorem, I find it useful for the students to show them that we are, in a sense, generalizing a fact they are already familiar with.

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This is quite an interesting application. Could you please elaborate on the part where you compare orders? Which elements did you compute the orders of? – Katie Dobbs Feb 21 '13 at 11:17
@KatieDobbs, thanks! I am comparing orders of quotient groups, or equivalently numbers of cosets. Using this second form, if $m \mid n$, so that $n \mathbf{Z}$ is contained in $m \mathbf{Z}$, then the number of cosets of $n \mathbf{Z}$ in $m \mathbf{Z}$ is $n/m$. – Andreas Caranti Feb 21 '13 at 11:32
@AndreasCaranti why aZ + bZ = gcd(a,b)Z? – Aha Nov 18 '14 at 2:29
@user2675516, it's a standard fact. Clearly $a \mathbf{Z}, b \mathbf{Z} \le \gcd(a, b) \mathbf{Z}$. Conversely, since you can write $\gcd(a, b) = a u + b v$, for appropriate $u, v \in \mathbf{Z}$, every multiple of $\gcd(a, b)$ is in $a \mathbf{Z} + b \mathbf{Z}$. – Andreas Caranti Nov 18 '14 at 12:18
@AndreasCaranti: I am having trouble translating form the statement of the third isomorphism theorem to what you wrote in your main equation line. Can you please elaborate? – Sandeep Silwal Oct 4 '15 at 20:38

Both are used a lot when studying normal series, for example derived, Fitting, lower/upper central, or composition series, and explicitly talking about its members.

If I'm trying to prove something about a Sylow $p$-subgroup of a quotient by a Sylow $q$-subgroup $(p\not= q)$, for example, $PQ/Q\cong P/(P\cap Q)$ is very useful because we know that $P\cap Q=1$. So we have $PQ/Q\cong P$, a intuitive result that the internal a Sylow subgroup is not affected by quotienting out by another unrelated Sylow subgroup. Plus, then in the rest of the proof, we know we can refer to the Sylow $p$-subgroups of $G/Q$ as $P$ instead of $PQ/Q$, which is notationally convenient.

Similarly, say we're looking at the series $Z_0=G$ and $Z_i/Z_{i-1}=Z(G/Z_{i-1})$. We know that $\frac{G/Z_{i-1}}{Z_i/Z_{i-1}}\cong G/Z_i$, which is a much easier way of talking about that group.

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It is true that the first isomorphism theorem is more commonly used than the second or third one. Zassenhaus Lemma uses the third isomorphism theorem. I can't think of a theorem that essentially uses the second isomorphism theorem, though it is useful in computations.

It should be noted that the second and third isomorphism theorems are direct consequences of the first, and in fact (somewhat philosophically) there is just one isomorphism theorem (the first one), the other two are corollaries.

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I think another well-known example of using that theorems maybe epitomize in Jordan-Hölder Theorem when we want to see that any two composition series of a given group are equivalent.

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Here and here are examples of questions I've answered with little more than the second isomorphism theorem.

The second one comes up a lot in modular arithmetic.

If you working in the integers modulo $16$? Maybe you'll want to reduce modulo $8$. Can you do that? And if so, do you get the integers modulo $8$?

$$ (\mathbb{Z} / 16\mathbb{Z}) / (8\mathbb{Z} / 16\mathbb{Z}) \cong \mathbb{Z} / 8\mathbb{Z} $$

Ah, phew!

The second isomorphism theorem often comes up when you want to do calculations with a quotient ring by lifting the problem to the original ring (e.g. because it is easier to work with, such as the integers or being a polynomial ring)

Now I'm working in the even numbers but I need to also work modulo $3$. How can I make sense of that? I suppose I need to mod out by the intersection of $3\mathbb{Z} \cap 2\mathbb{Z}$. What does that work out to?

$$2\mathbb{Z} / (2\mathbb{Z} \cap 3 \mathbb{Z}) \cong (2 \mathbb{Z} + 3 \mathbb{Z}) / 3 \mathbb{Z} \cong \mathbb{Z} / 3\mathbb{Z}$$

The third isomorphism theorem often comes up when you have several moduli to work with, and want to understand one ideal modulo the other.

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E.g., in studying solvable groups.

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Yes, definitely. I used both of these all over the place in my paper on soluble groups. – Tara B Feb 21 '13 at 10:45

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