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Proving the following sets are subgroups of an abelian group, G, is simple enough using the subgroup test, but I'm struggling to find counterexamples that show if G is NOT abelian, H is not necessarily a subgroup using the fact that G is not abelian.

$H_1=\{g \in G\ |\ g = g^{-1}\}$

$H_2=\{g \in G\ |\ g^n = e\}$ where n is a fixed integer.

The best I've come up with as a counterexample for $H_1$ is $G=D_3$ and $H_1=\{\text{the identity and the set of flips}\}$, but it's not a subgroup because its not closed, not because G is not abelian... or is it? I have nothing for $H_2$.

Shailesh
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LidStar
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  • For $n=2$ we have $H_1=H_2$. So you do have something for $H_2$. – Dietrich Burde Sep 15 '16 at 13:27
  • If $G$ is not abelian, $H_1$ might still be a subgroup, e.g., $G=A\times B$ where $A$ is abelian and $B$ is not and $B$ contains no elements of order 2. When $H_1$ is not a subgroup, that can only be because it's not closed. – Gerry Myerson Sep 15 '16 at 13:29

4 Answers4

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Take the free product of two cyclic groups, $G=C_n\ast C_m$, with $C_n=\langle a \mid a^n=e\rangle$ and $C_m=\langle b \mid b^m =e\rangle$. Then the product $ab$ has infinite order. So neither $H_1$ nor $H_2$ for any $n\in \mathbb{N}$ is a subgroup. We have used the fact, that the free product here is not abelian. More counterexamples are given in this question.

Community
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Dietrich Burde
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Consider the elements $a=(1,2,3,\ldots,n)$ and $b=(n,n+1,2n-1)$ of $S_{2n-1}$.

Then $a$ and $b$ are $n$-cycles and have order $n$, but $ab = (1,2,3,\ldots,2n-1)$ is a $(2n-1)$-cycle, so we have $a^n=b^n=1$ but $(ab)^n \ne 1$.

(By $ab$ I mean $b$ followed by $a$ - of course $ba$ is also a $(2n-1)$-cycle.)

Derek Holt
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Actually your choice of $D_3$ is correct for $H_1$. Recall that a subgroup of a group is a subset of the group's elements which satisfies three conditions: (1) it is closed under the group operation; (2) it contains the identity element; (3) it contains the inverse of each of its elements. Then a subset can fail to be a subgroup by failing to satisfy any of these conditions, so not being closed is enough.

And in this case, not being closed itself follows from being non-Abelian, which we can show by showing the contrapositive: If $H_1$ were closed, then $H_1$ would be Abelian. To see that: Let $a$, $b$ be in $H_1$. By closure, $ab$ is in $H_1$. By definition of $H_1$, $ab$ = $(ab)^{-1}$, or multiplying both sides by $ab$, $abab$=$e$. Multiply by $a$ on left, $b$ on right: $(aa)ba(bb)$=$a(e)b$, or by definition of $H_1$, $ba$=$ab$. SUMMARY: If $H_1$ is closed, any two elements commute --> Abelian.

PMar
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The example with $D_3$ for $H_1$ is good. For $H_2$ take $$ H_2=\{\sigma\in S_3:\sigma^2=\mathit{id}\} $$ where $S_3$ is the symmetric group on three objects and $\mathit{id}$ is the identity. Then $H_2$ has four elements: $$ H_2=\{\mathit{id},(12),(13),(23)\} $$ so it cannot be a subgroup; actually, it is the same as your $H_1$.

On the other hand $$ H_2'=\{\sigma\in S_3:\sigma^3=\mathit{id}\} $$ is a subgroup of $S_3$: $H_2'=\{\mathit{id},(123),(132)\}=\langle(123)\rangle$.

egreg
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