MTH3003 Weekly Problems 11

Original Documents: mth3003 weekly problem sheet 11.pdf / [[mth3003 weekly problem sheet 11 handwritten solutions.pdf]] / mth3003 weekly problem sheet 11 solutions.pdf

Vibes: The closing sheet, all on direct products and the classification of finite abelian groups. Two short structural proofs about external direct products, an embedding result for cyclic $p$-groups, a “list them all” computation, and a slick capstone showing abelian groups satisfy a converse to Lagrange.

Used Techniques:

• Quick subgroup test: identity, closure, inverses.
• Coordinate-swap isomorphism: $\theta (g,h)=(h,g)$ proves $G\times H\cong H\times G$.
• Fundamental Theorem of Finite Abelian Groups: classify by partitioning prime-power multiplicities.
• Subgroup notation $\lesssim$: write $H\lesssim G$ when $G$ has a subgroup isomorphic to $H$.

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11.1. The Direct Product is Commutative

Question

Let $G$ and $H$ be groups. Prove that $G\times H\cong H\times G$.

Hint: try the map $\theta :G\times H\to H\times G$ given by $\theta (g,h)=(h,g)$.

Define $\theta :G\times H\to H\times G$ by $\theta (g,h)=(h,g)$. Recall multiplication in a direct product is componentwise: $(g_1,h_1)(g_2,h_2)=(g_1{g}_2,h_1{h}_2)$.

Homomorphism.

$$\theta ((g_1,h_1)(g_2,h_2))=\theta (g_1{g}_2,h_1{h}_2)=(h_1{h}_2,g_1{g}_2),$$

and on the other side

$$\theta (g_1,h_1)\theta (g_2,h_2)=(h_1,g_1)(h_2,g_2)=(h_1{h}_2,g_1{g}_2).$$

These agree, so $\theta$ is a Homomorphism.

Injective. If $\theta (g_1,h_1)=\theta (g_2,h_2)$ then $(h_1,g_1)=(h_2,g_2)$, so $h_1=h_2$ and $g_1=g_2$, hence $(g_1,h_1)=(g_2,h_2)$.

Surjective. Any $(h,g)\in H\times G$ equals $\theta (g,h)$.

So $\theta$ is an Isomorphism and $G\times H\cong H\times G$. $■$

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11.2. Direct Product of Subgroups is a Subgroup

Question

Let $G$ and $H$ be groups with $K\le G$ and $J\le H$. Prove that $K\times J\le G\times H$.

Use the quick subgroup test on the subset $K\times J\subseteq G\times H$.

Identity. Since $K\le G$ and $J\le H$, we have $e_G\in K$ and $e_H\in J$, so $(e_G,e_H)\in K\times J$ - and this is the identity of $G\times H$.

Closure and inverses. Take $(g_1,h_1),(g_2,h_2)\in K\times J$. Then

$$(g_1,h_1)(g_2,h_2{)}^{-1}=(g_1{g}_2^{-1},h_1{h}_2^{-1}).$$

Because $K$ is a group, $g_1,g_2\in K\Rightarrow g_1{g}_2^{-1}\in K$; because $J$ is a group, $h_1,h_2\in J\Rightarrow h_1{h}_2^{-1}\in J$. Hence $(g_1{g}_2^{-1},h_1{h}_2^{-1})\in K\times J$.

So $K\times J$ is non-empty and closed under products-with-inverses, giving $K\times J\le G\times H$. $■$

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11.3. Cyclic $p$-Groups Embed in Larger Cyclic $p$-Groups

Question

Let $p$ be a prime and $r,s\ge 0$ integers with $s\le r$. Prove that $C_{p^s}\lesssim C_{p^r}$, i.e. $C_{p^r}$ has a subgroup isomorphic to $C_{p^s}$.

Hint: find an element $h\in C_{p^r}$ of order $p^s$, then $H=⟨h⟩\cong C_{p^s}$.

If $s=0$ then $C_{p^s}=\{e\}\lesssim C_{p^r}$ trivially, and if $s=r$ then $C_{p^s}=C_{p^r}$, so assume $1\le s<r$.

Write $C_{p^r}=⟨\tau ⟩$ where $\tau =(12\cdots p^r)$ has order $p^r$. Set $h={\tau }^{p^{r-s}}$ and claim $o(h)=p^s$.

Order at most $p^s$.

$$h^{p^s}=({\tau }^{p^{r-s}}{)}^{p^s}={\tau }^{p^{r-s}\cdot p^s}={\tau }^{p^r}=e,$$

so $o(h)\le p^s$.

Order at least $p^s$. For $0<\ell <p^s$ we have $h^{\ell }={\tau }^{p^{r-s}\ell }$, and $p^{r-s}\ell <p^{r-s}{p}^s=p^r=o(\tau )$, so $h^{\ell }\ne e$.

Hence $o(h)=p^s$, and $H=⟨h⟩\le C_{p^r}$ is cyclic of order $p^s$, so $H\cong C_{p^s}$. Therefore $C_{p^s}\lesssim C_{p^r}$. $■$

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11.4. All Abelian Groups of Order 72

Question

Find, up to isomorphism, all abelian groups of order $72$.

Factorise $72=2^3\cdot 3^2$. By the Fundamental Theorem of Finite Abelian Groups, every abelian group of order $72$ is a direct product of cyclic prime-power groups, and distinct decompositions give non-isomorphic groups. So count independently the partitions of the exponent $3$ on the prime $2$ and the exponent $2$ on the prime $3$.

• Partitions of $3$ (for the $2$-part): $(3),(2,1),(1,1,1)$ - three of them.
• Partitions of $2$ (for the $3$-part): $(2),(1,1)$ - two of them.

This gives $3\times 2=6$ groups.

$2$-part	$3$-part	Decomposition	Group
$(3)$	$(2)$	$2^3\cdot 3^2$	$C_{2^3}\times C_{3^2}$
$(3)$	$(1,1)$	$2^3\cdot 3\cdot 3$	$C_{2^3}\times C_3\times C_3$
$(2,1)$	$(2)$	$2^2\cdot 2\cdot 3^2$	$C_{2^2}\times C_2\times C_{3^2}$
$(2,1)$	$(1,1)$	$2^2\cdot 2\cdot 3\cdot 3$	$C_{2^2}\times C_2\times C_3\times C_3$
$(1,1,1)$	$(2)$	$2\cdot 2\cdot 2\cdot 3^2$	$C_2\times C_2\times C_2\times C_{3^2}$
$(1,1,1)$	$(1,1)$	$2\cdot 2\cdot 2\cdot 3\cdot 3$	$C_2\times C_2\times C_2\times C_3\times C_3$

$$\boxed{C_8\times C_9,C_8\times C_3^2,C_4\times C_2\times C_9,C_4\times C_2\times C_3^2,C_2^3\times C_9,C_2^3\times C_3^2.}$$

So there are $6$ abelian groups of order $72$ up to isomorphism. $■$

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11.5. Abelian Groups Satisfy the Converse of Lagrange

Question

Is the following true? Give a proof or counterexample. If $G$ is a finite abelian group of order $m$ and $k\mid m$, then $G$ has a subgroup of order $k$.

Hint: a tricky one - use the Fundamental Theorem together with Problems 11.2 and 11.3.

True for abelian groups (it fails in general - $A_4$ has order $12$ but no subgroup of order $6$, but $A_4$ is not abelian).

By the Fundamental Theorem of Finite Abelian Groups there are (not necessarily distinct) primes $p_1,\dots ,p_n$ and natural numbers $e_1,\dots ,e_n$ with $p_1^{e_1}\cdots p_n^{e_n}=m$ and

$$G\cong C_{p_1^{e_1}}\times \cdots \times C_{p_n^{e_n}}.$$

Since $k\mid m$, the divisor $k$ uses only these primes, so

$$k=p_1^{d_1}\cdots p_n^{d_n},0\le d_i\le e_i.$$

For each $i$, Problem 11.3 gives $C_{p_i^{d_i}}\lesssim C_{p_i^{e_i}}$ - choose a subgroup $H_i\le C_{p_i^{e_i}}$ with $H_i\cong C_{p_i^{d_i}}$. By Problem 11.2 (applied repeatedly), the product

$$H_1\times \cdots \times H_n\le C_{p_1^{e_1}}\times \cdots \times C_{p_n^{e_n}}\cong G$$

is a subgroup, and its order is $p_1^{d_1}\cdots p_n^{d_n}=k$. Hence $G$ has a subgroup of order $k$. $■$

Warning

This converse to Lagrange’s theorem is special to abelian groups (more generally, nilpotent groups). For arbitrary finite groups it is false: $|A_4|=12$ but $A_4$ has no subgroup of order $6$.