Modern Algebra 1 — Homework 1 Solutions

Spring 2026  |  Due: April 7
KIM Süngsoo  ·  2022-14643
Problem 1
Claim. If $G$ is a finite group with identity $e$ and $|G|$ is even, then there exists $a \neq e$ in $G$ such that $a * a = e$.

Proof.

Define the inversion map $\iota : G \to G$ by $\iota(a) = a^{-1}$. Since $(a^{-1})^{-1} = a$ for every $a \in G$, this map is an involution: $\iota^2 = \id_G$.

The involution $\iota$ partitions $G$ into orbits under the action of $\langle \iota \rangle \cong \Z_2$. Each orbit has size 1 or 2:

G F = Fix(ι) e a ⋯ |F| is even Free orbits (size 2) b b⁻¹ c c⁻¹ contributes even count
Figure 1. Partition of $G$ under the inversion involution $\iota: a \mapsto a^{-1}$.

Since every element lies in exactly one orbit, we have the orbit-counting equation: $$|G| \;=\; |F| \;+\; 2k$$ for some non-negative integer $k$ (the number of free orbits). Because $|G|$ is even and $2k$ is even, $|F|$ must also be even.

Now $e \in F$ (since $e^2 = e$), so $|F| \geq 1$. But $|F|$ is even, which forces $|F| \geq 2$. Therefore there exists some $a \in F$ with $a \neq e$, meaning $a^2 = e$.

$\blacksquare$
Remark (categorical perspective). The existence of an element of order $2$ is equivalent to the existence of a non-trivial group homomorphism $\Z_2 \to G$. In the category $\Grp$, this is encoded by the following commutative diagram, where $\varphi(\bar{1}) = a$ with $a^2 = e$: $$\begin{CD} \Z_2 @>{\exists\;\varphi \,\neq\, 0}>> G \\ @V{\id}VV @VV{\iota}V \\ \Z_2 @>>{\varphi}> G \end{CD}$$ The diagram commutes because $\iota(\varphi(\bar{1})) = a^{-1} = a = \varphi(\bar{1})$, i.e. $\varphi$ maps into the fixed-point set of $\iota$. This problem is essentially the $p = 2$ case of Cauchy's theorem.
Problem 2
Claim. The number of cycles $\sigma$ in $S_5$ such that $\sigma^2$ is also a cycle (counting the identity as a cycle) is $\mathbf{55}$.

Proof.

Definition. A $k$-cycle in $S_n$ is a permutation $\sigma$ of the form $\sigma = (a_1 \; a_2 \; \cdots \; a_k)$, meaning $\sigma(a_i) = a_{i+1}$ for $1 \leq i \leq k-1$, $\;\sigma(a_k) = a_1$, and $\sigma$ fixes all elements outside $\{a_1, \ldots, a_k\}$. A $1$-cycle is the identity, and a $2$-cycle is a transposition.

Lemma. Let $\sigma = (a_1 \; a_2 \; \cdots \; a_k)$ be a $k$-cycle. Then $\sigma^2$ consists of exactly $d = \gcd(2, k)$ disjoint cycles, each of length $k/d$. In particular, $\sigma^2$ is a cycle if and only if $\gcd(2, k) = 1$ (i.e., $k$ is odd) or $k \leq 2$.

Proof of Lemma.

By definition, $\sigma(a_i) = a_{i+1 \pmod{k}}$, so $\sigma^2(a_i) = a_{i+2 \pmod{k}}$. Thus $\sigma^2$ acts on the index set $\Z / k\Z$ by the map $i \mapsto i + 2$. The orbit of index $i$ under this map is $\{i, \; i+2, \; i+4, \; \ldots\}$ taken modulo $k$.

The size of this orbit equals the order of $\bar{2}$ in $\Z / k\Z$, which is $k / \gcd(2, k)$. The number of distinct orbits is $\gcd(2, k)$, since $|\Z / k\Z|$ = (number of orbits) $\times$ (orbit size). Concretely:

  • $k$ odd ($\gcd(2,k) = 1$): the single orbit $\{0, 2, 4, \ldots, k-1, 1, 3, \ldots\}$ visits every index, so $\sigma^2$ is a $k$-cycle.
  • $k$ even ($\gcd(2,k) = 2$): there are two orbits — the even indices $\{0, 2, 4, \ldots, k-2\}$ and the odd indices $\{1, 3, 5, \ldots, k-1\}$ — each of size $k/2$. So $\sigma^2$ is a product of two disjoint $(k/2)$-cycles.

Hence $\sigma^2$ is a single cycle precisely when $k$ is odd (giving a $k$-cycle) or $k \leq 2$ (where $\sigma^2 = \id$, which we count as a cycle).

$\blacksquare$

Applying the lemma, $\sigma^2$ is a cycle for $k \in \{1, 2, 3, 5\}$ and fails to be a cycle for $k = 4$ (where $\sigma^2$ splits into two transpositions).

We now enumerate each case in $S_5$. Recall that the number of $k$-cycles in $S_n$ is $\displaystyle\binom{n}{k} \cdot (k-1)!\,$.

Length $k$ $\sigma^2$ structure $\sigma^2$ a cycle? Count in $S_5$
$1$ (identity) identity Yes $1$
$2$ (transposition) identity Yes $\displaystyle\binom{5}{2}\cdot 1! = 10$
$3$ $3$-cycle Yes $\displaystyle\binom{5}{3}\cdot 2! = 20$
$4$ product of two $2$-cycles No $0$ (excluded)
$5$ $5$-cycle Yes $\displaystyle\binom{5}{5}\cdot 4! = 24$

Verification for $k = 3$:

Let $\sigma = (1\;2\;3)$. Then $\sigma^2(1) = 3$, $\sigma^2(3) = 2$, $\sigma^2(2) = 1$, giving $\sigma^2 = (1\;3\;2)$, which is indeed a $3$-cycle.

Verification for $k = 4$:

Let $\sigma = (1\;2\;3\;4)$. Then $\sigma^2(1) = 3$, $\sigma^2(3) = 1$, $\sigma^2(2) = 4$, $\sigma^2(4) = 2$, giving $\sigma^2 = (1\;3)(2\;4)$, which is not a single cycle.

Verification for $k = 5$:

Let $\sigma = (1\;2\;3\;4\;5)$. Then: $$\sigma^2(1) = 3, \quad \sigma^2(3) = 5, \quad \sigma^2(5) = 2, \quad \sigma^2(2) = 4, \quad \sigma^2(4) = 1,$$ giving $\sigma^2 = (1\;3\;5\;2\;4)$, which is indeed a $5$-cycle.

1 2 3 4 5 σ = (1 2 3 4 5) σ² = (1 3 5 2 4)
Figure 2. Pentagon vs. pentagram: $\sigma$ traverses edges (blue, connected), $\sigma^2$ traverses the star (red, still connected — a single cycle).

Total: $$1 + 10 + 20 + 0 + 24 \;=\; \boxed{55}$$

$\blacksquare$
Remark (generalization). The lemma generalizes: for a $k$-cycle $\sigma$ and any positive integer $m$, the permutation $\sigma^m$ consists of $\gcd(m, k)$ disjoint cycles each of length $k / \gcd(m, k)$. In particular, $\sigma^m$ is again a cycle if and only if $\gcd(m, k) = 1$ or $k \mid m$ (where $\sigma^m = \id$). The proof is identical — $\sigma^m$ acts on indices by $i \mapsto i + m \pmod{k}$, and the orbit size of this map is $k / \gcd(m, k)$.

For the present problem ($m = 2$), this reduces to: $\sigma^2$ is a cycle iff $k$ is odd (or $k \leq 2$ trivially). One could therefore bypass the case analysis entirely and write: the qualifying cycles are precisely those of odd length, plus the transpositions. This gives $$\underbrace{1}_{\id} \;+\; \underbrace{\binom{5}{2} \cdot 1!}_{k=2} \;+\; \underbrace{\binom{5}{3} \cdot 2!}_{k=3} \;+\; \underbrace{\binom{5}{5} \cdot 4!}_{k=5} \;=\; 1 + 10 + 20 + 24 = 55.$$
Problem 3
Theorem (Fundamental Theorem of Finitely Generated Abelian Groups). Every finitely generated abelian group $G$ is isomorphic to $$G \;\cong\; \Z^{\,r} \times \Z_{n_1} \times \Z_{n_2} \times \cdots \times \Z_{n_s}$$ where $r \geq 0$ and $n_1 \mid n_2 \mid \cdots \mid n_s$ with each $n_i > 1$. The integer $r$ (the free rank) and the integers $n_1, \ldots, n_s$ (the invariant factors) are uniquely determined by $G$.

Equivalently, $G$ can be decomposed into elementary divisors (prime-power cyclic factors), and the two decompositions determine each other.
Claim. The invariant factor decomposition of $\Z_6 \times \Z_{12} \times \Z_{20}$ is $\;\Z_2 \times \Z_{12} \times \Z_{60}$,
i.e., $m_1 = 2$, $m_2 = 12$, $m_3 = 60$.

Proof.

Step 1: Primary (elementary divisor) decomposition via the Chinese Remainder Theorem.

We use the following form of the CRT for cyclic groups:

Chinese Remainder Theorem. If $n = p_1^{a_1} p_2^{a_2} \cdots p_s^{a_s}$ is the prime factorization of $n$, then $$\Z_n \;\cong\; \Z_{p_1^{a_1}} \times \Z_{p_2^{a_2}} \times \cdots \times \Z_{p_s^{a_s}}.$$ The isomorphism is given by the map $\;[x]_n \;\mapsto\; \bigl([x]_{p_1^{a_1}},\; [x]_{p_2^{a_2}},\; \ldots,\; [x]_{p_s^{a_s}}\bigr)$, which is well-defined because $p_i^{a_i} \mid n$ for each $i$. This is an isomorphism precisely because $\gcd(p_i^{a_i}, p_j^{a_j}) = 1$ for $i \neq j$.

We apply this to each factor. First, the prime factorizations: $$6 = 2 \cdot 3, \qquad 12 = 2^2 \cdot 3, \qquad 20 = 2^2 \cdot 5.$$

Applying the CRT to each:

Now substitute into the original product and rearrange by prime:

$$\Z_6 \times \Z_{12} \times \Z_{20} \;\cong\; (\Z_2 \times \Z_3) \times (\Z_4 \times \Z_3) \times (\Z_4 \times \Z_5)$$ $$=\; \underbrace{\Z_2 \times \Z_4 \times \Z_4}_{2\text{-primary component } G_2} \;\times\; \underbrace{\Z_3 \times \Z_3}_{3\text{-primary component } G_3} \;\times\; \underbrace{\Z_5}_{5\text{-primary component } G_5}.$$

This uses associativity and commutativity of the direct product (valid in the category of abelian groups). The elementary divisors of the group are exactly these prime-power factors: $2, 4, 4, 3, 3, 5$.

The overall decomposition pipeline is:

$$\begin{CD} \Z_6 \times \Z_{12} \times \Z_{20} @>{\text{CRT per factor}}>> G_2 \times G_3 \times G_5 @>{\text{recombine columns}}>> \Z_{m_1} \times \Z_{m_2} \times \Z_{m_3} \end{CD}$$

Step 2: From elementary divisors to invariant factors.

The passage from elementary divisors to invariant factors works as follows. For each prime $p$, list the $p$-power elementary divisors in non-decreasing order. Since different primes may contribute different numbers of factors, pad the shorter lists on the left with $1$'s so all lists reach the same length $r$ (the maximum). Then the $j$-th invariant factor $m_j$ is the product of the $j$-th entries across all primes. Because we padded on the left and each column's entries are coprime (one entry per prime), the CRT guarantees $\Z_{m_j} \cong \prod_p \Z_{p\text{-entry in column }j}$, and the divisibility $m_j \mid m_{j+1}$ follows from the non-decreasing arrangement within each row.

In our case, $r = 3$ (the $2$-primary component has $3$ factors, the most of any prime):

Prime $p$ Column 1 (smallest) Column 2 Column 3 (largest)
$p = 2$:   $\Z_2 \times \Z_4 \times \Z_4$ $2^1 = 2$ $2^2 = 4$ $2^2 = 4$
$p = 3$:   $\Z_3 \times \Z_3$ $3^0 = 1$ (pad) $3^1 = 3$ $3^1 = 3$
$p = 5$:   $\Z_5$ $5^0 = 1$ (pad) $5^0 = 1$ (pad) $5^1 = 5$
Invariant factor $m_j = \prod_p$ $m_1 = 2 \cdot 1 \cdot 1 = 2$ $m_2 = 4 \cdot 3 \cdot 1 = 12$ $m_3 = 4 \cdot 3 \cdot 5 = 60$

Each invariant factor recombines via CRT: $$m_1 = 2, \qquad m_2 = 4 \cdot 3 = 12, \qquad m_3 = 4 \cdot 3 \cdot 5 = 60.$$ Equivalently: $\;\Z_2 \cong \Z_2$, $\;\Z_{12} \cong \Z_4 \times \Z_3$, $\;\Z_{60} \cong \Z_4 \times \Z_3 \times \Z_5$.

Step 3: Verification.

We verify the divisibility chain: $m_1 \mid m_2$ since $2 \mid 12$, and $m_2 \mid m_3$ since $12 \mid 60$, as required. The product $m_1 \cdot m_2 \cdot m_3 = 2 \cdot 12 \cdot 60 = 1440 = 6 \cdot 12 \cdot 20 = |G|$ confirms the order is preserved.

As a consistency check, we reverse the process and recover the elementary divisors: $$\Z_2 \times \Z_{12} \times \Z_{60} \;\overset{\text{CRT}}{\cong}\; \Z_2 \times (\Z_4 \times \Z_3) \times (\Z_4 \times \Z_3 \times \Z_5) \;=\; \Z_2 \times \Z_4 \times \Z_4 \times \Z_3 \times \Z_3 \times \Z_5$$ which agrees with the primary decomposition obtained in Step 1. By uniqueness of the invariant factor decomposition (guaranteed by the Fundamental Theorem), this is the only valid decomposition with $m_i > 1$ and $m_i \mid m_{i+1}$.

Therefore: $\;\boxed{m_1 = 2, \quad m_2 = 12, \quad m_3 = 60.}$

$\blacksquare$
Problem 4
Claim. Let $H, K$ be subgroups of a group $G$ satisfying:
(a) every element of $G$ is of the form $hk$ for some $h \in H$, $k \in K$;
(b) $hk = kh$ for all $h \in H$, $k \in K$;
(c) $H \cap K = \{e\}$.
Then $G \cong H \times K$.

Proof.

Define the map

$$\varphi : H \times K \longrightarrow G, \qquad \varphi(h, k) = hk.$$

The key structures are captured by the following commutative diagram, where $\iota_H : H \hookrightarrow G$ and $\iota_K : K \hookrightarrow G$ are the inclusion homomorphisms and $j_H(h) = (h, e)$, $j_K(k) = (e, k)$ are the canonical embeddings:

$$\begin{CD} H @>{j_H}>> H \times K @<{j_K}<< K \\ @| @VV{\varphi}V @| \\ H @>>{\iota_H}> G @<<{\iota_K}< K \end{CD}$$

We show that $\varphi$ is an isomorphism by verifying the three required properties.

Step 1: $\varphi$ is a homomorphism.

Let $(h_1, k_1),\, (h_2, k_2) \in H \times K$. On one hand: $$\varphi\bigl((h_1, k_1)(h_2, k_2)\bigr) = \varphi(h_1 h_2,\, k_1 k_2) = h_1 h_2 \cdot k_1 k_2.$$ On the other hand: $$\varphi(h_1, k_1) \cdot \varphi(h_2, k_2) = h_1 k_1 \cdot h_2 k_2 = h_1 \underbrace{(k_1 h_2)}_{\overset{(b)}{=}\; h_2 k_1} k_2 = h_1 h_2 \cdot k_1 k_2.$$ The two expressions agree, so $\varphi$ is a homomorphism.

Step 2: $\varphi$ is surjective.

By condition (a), every $g \in G$ can be written as $g = hk$ for some $h \in H$, $k \in K$. Thus $g = \varphi(h, k)$, so $\varphi$ is surjective.

Step 3: $\varphi$ is injective.

Suppose $\varphi(h, k) = e_G$, i.e., $hk = e$. Then: $$h = k^{-1}.$$ Now $h \in H$ and $k^{-1} \in K$ (since $K$ is a subgroup, closed under inverses). Therefore: $$h = k^{-1} \in H \cap K \;\overset{(c)}{=}\; \{e\}.$$ This forces $h = e$ and $k = e$, so $\ker \varphi = \{(e, e)\}$, meaning $\varphi$ is injective.

Conclusion.

Since $\varphi$ is a bijective homomorphism, it is an isomorphism: $$\boxed{G \;\cong\; H \times K.}$$

$\blacksquare$
Remark (universal property of the direct product in $\Grp$). In the category $\Grp$, the direct product $H \times K$ satisfies the following universal property: for any group $X$ with homomorphisms $f : X \to H$ and $g : X \to K$, there exists a unique homomorphism $(f, g) : X \to H \times K$ making the diagram commute. $$\begin{CD} @. X \\ @. @V{(f,\,g)}VV \\ H @<<{\pi_H}< H \times K @>>{\pi_K}> K \\ @. @VV{\varphi\;\cong}V \\ @. G \end{CD}$$ Conditions (a)–(c) assert precisely that $G$ realizes this universal property internally: the multiplication map $\varphi(h,k) = hk$ is the canonical comparison morphism, and the three conditions force it to be an isomorphism. This is a special case of the general principle that a group is an internal direct product if and only if the canonical map from the corresponding external direct product is an isomorphism.
Remark (alternative proof via uniqueness of representation). One can also prove the isomorphism by working from $G$ to $H \times K$, rather than the other direction. By (a), every $g \in G$ admits a factorization $g = hk$. We first show this factorization is unique: if $h_1 k_1 = h_2 k_2$ with $h_i \in H$, $k_i \in K$, then $h_2^{-1} h_1 = k_2 k_1^{-1}$. The left side lies in $H$ and the right in $K$, so both lie in $H \cap K = \{e\}$ by (c), giving $h_1 = h_2$ and $k_1 = k_2$.

Now define $\psi : G \to H \times K$ by $\psi(g) = (h, k)$ where $g = hk$ is the unique factorization. This is well-defined by uniqueness, and is evidently the inverse of $\varphi$. To see that $\psi$ is a homomorphism (equivalently, that $\varphi$ is), one uses condition (b) exactly as in Step 1 above. This approach has the advantage of making the bijection between $G$ and $H \times K$ manifest before verifying the algebraic structure.
— End of Solutions —