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I am trying to solve the below problem.

Below are several examples of groups. Among these, determine which are isomorphic to each other or to subgroups of other groups.

  1. The trivial group $\{e\}$.
  2. Number systems with addition: $\left(\mathbb{Z}, +\right)$, $\left(\mathbb{Q}, +\right)$, $\left(\mathbb{R}, +\right)$, $\left(\mathbb{C}, +\right)$.
  3. $\{0,1\}$ with addition mod 2; $\{\pm 1\}$ with multiplication.
  4. $\mathbb{Z}/n = \{0,1, \ldots, n-1\}$ with addition mod $n$ and $\mathbb{R}/\mathbb{Z}$, which is the set $[0,1)$ with the addition law $(a,b) \mapsto$ $a+b$ if $a + b < 1$ and $a + b - 1$ otherwise.
  5. Non-zero numbers with multiplication: $\mathbb{Q}^{\star}$, $\mathbb{R}^{\star}$, $\mathbb{C}^{\star}$, and the unit circle $S^1 \subset \mathbb{C}^{\star}$.
  6. Permutations of a set $A$, $\mathrm{Perm}(A)$, and symmetric group on $n$ letters $S_n$.
  7. $\mathrm{GL}_n (F)$ and $\mathrm{SL}_n (F)$ for some field $F$.
  8. Direct products and direct sums of finitely or infinitely many groups.

Here is my attempt. I am focusing more on finding the isomorphisms and understanding them intuitively than directly proving them, which I'm going to do later on, but don't have too much trouble with.

As $\mathbb{Q} \cong \mathbb{Z}$ as sets, $\left(\mathbb{Z}, +\right) \cong \left(\mathbb{Q}, +\right)$. Since there exists a natural embedding $\mathbb{R} \hookrightarrow \mathbb{C}$ sending $x \mapsto x + 0i$, we can identify $\left(\mathbb{R}, +\right)$ with a subgroup of $\left(\mathbb{C}, +\right)$, namely the real axis under addition. Similarly, we can embed $\mathbb{Q}$ in $\mathbb{R}$, so we can identify $(\mathbb{Q}, +)$ with a subgroup of $(\mathbb{R}, +)$.

We can proceed exactly analagously for multiplication. Namely, we can identify $\mathbb{Q}^{\star}$ with a subgroup of $\mathbb{R}^{\star}$, $\mathbb{R}^{\star}$ with a subgroup of $\mathbb{C}^{\star}$, and $S^1$ with a subgroup of $\mathbb{C}^{\star}$.

Permutations of the set $A = \{1, \ldots, n\}$ and $S_n$ are surely isomorphism, and $\mathrm{SL}_n (F)$ can be identified with a subgroup of $\mathrm{GL}_n (F)$.

Direct sums can be identified with a subgroups of direct products, holding the groups $G_i$ fixed.

I'm surely missing a few. There is a hint to consider $\mathbb{Z}/n$, $\mathbb{C}^{\star}$, and $GL_2 (\mathbb{R})$, but I can't think of any.

David A. Craven
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Brad G.
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    If $A \simeq B$ as sets, it is not necessarily true that $(A, +) \simeq (B, +)$. – Trebor Sep 01 '21 at 03:57
  • The simplest example is $\text{Klein's four group} \not \simeq \text{Cyclic four group}$. – Trebor Sep 01 '21 at 03:58
  • The interesting thing is, direct sums are isomorphic to direct products with finitely many groups :P – Trebor Sep 01 '21 at 04:08
  • $(\mathbb Z,+)$ and $(\mathbb Q,+)$ are not isomorphic as groups. The easiest way to see this is to note that for any $a\in\mathbb Q$, there exists $b\in\mathbb Q$ such that $b+b=a$, but that's not true in $\mathbb Z$, so there's no homomorphism from $\mathbb Q$ into $\mathbb Z$. – Michael Hartley Sep 01 '21 at 04:24
  • Perhaps you should also prove that $S^1$ is isomorphic to $\mathbb{R}/\mathbb{Z}$ and that $\mathbb{R}^$ is not isomorphic to $\mathbb{C}^$ (this is not because one is a proper subgroup of the other). Moreover, $\mathbb{Z}/n$ can be embedded in to $\mathbb{C}^$ and $\mathbb{C}^$ embeds into $GL_2(\mathbb{R})$ (this is one way to define complex numbers). – Brauer Suzuki Sep 01 '21 at 05:07
  • In 3 they are. In 6 if and only if $card(A)=n$. –  Sep 01 '21 at 05:30
  • Also $\mathbb{R}$ and $\mathbb{C}$ are isomorphic, see https://math.stackexchange.com/questions/925706/is-it-true-that-group-of-real-numbers-under-addition-isomorphic-to-group-of-comp – Brauer Suzuki Sep 01 '21 at 06:20
  • @Trebor, what do you mean by "$A \simeq B$ as sets"? –  Sep 01 '21 at 07:13
  • @CAB It's just what the OP means by $A\cong B$ I assume. There exists an isomorphism between $A$ and $B$. – Trebor Sep 01 '21 at 07:48
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    Don't forget $S_3 \cong \operatorname{GL}_2(2) = \operatorname{SL}_2(2)$, and $\operatorname{SL}_4(2)$ is isomorphic to $A_8$ which is a subgroup of $S_8$ (this question taken literally is completely ridiculous for a beginner). – Sean Eberhard Sep 01 '21 at 09:39
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    I think these are too many questions packed in a single one. – Dietrich Burde Sep 01 '21 at 12:18

1 Answers1

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Let me summarize some of the non-trivial (non)isomorphisms:

  • $(\mathbb{Z},+)\not\cong(\mathbb{Q},+)$, since the former is cyclic and the latter not even finitely generated.
  • $(\mathbb{Q},+)\not\cong(\mathbb{R},+)$, since the cardinality is different.
  • $(\mathbb{R},+)\cong(\mathbb{C},+)$, since those are $\mathbb{Q}$-vectorspaces of the same dimension (assuming axiom of choice).
  • $\mathbb{Z}^*\cong\mathbb{Z}/2\mathbb{Z}$ where $\mathbb{Z}^*=\{1,-1\}$ denotes the group of units of $\mathbb{Z}$.
  • $\mathbb{Q}^*\cong\mathbb{Z}^*\times\bigoplus_{n=1}^\infty\mathbb{Z}$ via the prime factorization $\pm 2^{a_1}3^{a_2}5^{a_3}\ldots\mapsto (\pm1,a_1,a_2,\ldots)$.
  • $\mathbb{R}^*\not\cong\mathbb{C}^*$, since the latter has elements of order $3$ (roots of unity).
  • $\mathbb{R}/\mathbb{Z}\cong S^1$ by the homomorphism $r\mapsto e^{2\pi r}$ with kernel $\mathbb{Z}$.
  • $\mathrm{SL}_n(F)\not\cong \mathrm{GL}_n(F)$ unless $|F|=2$. Almost always, $SL_n(F)$ is perfect and $GL_n(F)$ is not.
  • $\mathrm{GL}_n(F)\cong\mathrm{GL}_m(F')$ if and only if $n=m$ and $F\cong F'$. See Is there a simple way of proving that $\text{GL}_n(R) \not\cong \text{GL}_m(R)$?.
Brauer Suzuki
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    Don't forget $\mathbb{Q}^\times\cong \mathbb Z/2\mathbb Z\oplus \bigoplus_n \mathbb Z$, and if $F$ has no non-trivial $n$th roots of unity, $\mathrm{GL}_n(F)\cong \mathrm{SL}_n(F)\times F^\times$. – David A. Craven Sep 01 '21 at 12:02