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The task is as follows: Let $G$ be a finite group.

a) If $|G|$ is divisible by $2$, then there is an element of $G$ with order $2$.

b) If $G$ is abelian group and $|G|$ is divisible by ${2}^{n}$ ($n \in \mathbb{N}$), then there is a subgroup of $G$ with order ${2}^{n}$ .

I have already solved part a). But I need help to prove part b).

The fundamental theorem of abelian groups is not known yet.

It is advised to prove it via induction:

Induction start: $n=0$. Then the trivial subgroup $\{e\}$ is a subgroup of $G$.

Induction step: Let b) be true for an $n \in \mathbb{N}$ and let $G$ be a finite group with $|G|$ divisible by ${2}^{n+1}$. There is a subgroup $U$ with order ${2}^{n}$. Because $G$ is abelian $U$ is a normal subgroup of $G$ and therefore $G/U$ is a group.

Part a) says that there is an element $\overline{b} \in G/U$ with $\operatorname{ord}(b) = 2$.

That means: ${\overline{b}}^{2} = U$ (Operation in $G/U$)

So $\langle${ c : $\overline{b}=\overline{c}$}$\rangle$ (Operation in $G$) is a subgroup of $G$ with |$\langle${ c : $\overline{b}=\overline{c}$}$\rangle$| = $2*{2}^{n}$ = ${2}^{n+1}$

Can someone please review my proof?

calculatormathematical
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StudentSeekingHelp
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  • Use $\langle a\rangle$ for $\langle a\rangle$. – Shaun Oct 28 '23 at 12:53
  • Thank you, I corrected it. – StudentSeekingHelp Oct 28 '23 at 13:56
  • Can my question please be reopened? I am not allowed to use the fundamental theorem of abelian groups to prove b). It's an exercise on factor groups. – StudentSeekingHelp Oct 28 '23 at 14:55
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    It won't do. You assert that there is an element of order $2^n$ and that may not be true. You are looking for a subgroup of order $2^n$, not an element of order $2^n$. – ancient mathematician Oct 28 '23 at 15:25
  • Then let's assert there is a subgroup U with order ${2}^{n}$. Then G/U is still a group. Part a) says that there is an element $\overline{b} \in G/U$ with ord(b) = 2. That means: $ {\overline{b} }^{2}=U $. The subgroup generated by the elements of $\overline{b}$ does have ${2}^{n+1}$ elements. Is this proof correct? – StudentSeekingHelp Oct 28 '23 at 16:21
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    You need to be careful, but yeah $U \cup bU$ is a subgroup with $2^{n+1}$ elements, though you still need to show it’s a subgroup (or do something clever like note that it’s the pre image of the projection of the group generated by $bU$) – Eric Oct 28 '23 at 18:30

2 Answers2

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The notation $\langle c \in \overline{b} \rangle$ does not mean anything meaningful.

Instead, use a) to find an element $g \in G$ of order $2$. Since $G$ is abelian, $U := \langle g \rangle \subseteq G$ is a normal subgroup. Now apply the induction hypothesis to the group $G/U$, which has order $\mathrm{ord}(G)/2$.

You can use the correspondence theorem for subgroups.

Martin Brandenburg
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Nice exercise, StudentSeekingHelp. Your approach for an inductive proof is correct. Let's go through your induction step:

Let b) be true for some n∊ℕ₀ and let G be finite abelian group, |G| being a multiple of 2ⁿ⁺¹. Then there is a subgroup S of G with |S|=2ⁿ. Since S is normal subgroup of G G/S is group. |G/S|=|G|/|S| is even. Thus, according to part a), there's an x∊G such that x̅∊G/S has order 2. x itself needn't have order 2 (in G), but in any case not x∊S.

Let X≔{x̅, 1̅}, let π denote the natural epimorphism from G onto G/S. Since ord(x̅)=2 X is a subgroup of G/S. Thus its preimage π⁻¹(X)≕P is a subgroup of G. Note that P=x·S ∪ 1·S and x·S ∩ 1·S=∅, the latter since not x∊S. Thus |P|=2·|S|=2ⁿ⁺¹.

Peterש
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