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There is a lemma that says t if a group $G$ has no proper nontrivial subgroups, then $G$ is cyclic. And here is the proof of the lemma:

Suppose $G$ has no proper nontrivial subgroups. Take an element $a$ in $G$ for which $a$ is not equal to $e$. Consider the cyclic subgroup $\langle a \rangle$. This subgroup contains at least e and a, so it is not trivial. But G has no proper subgroups, so it must be that $\langle a \rangle = G$. Thus $G$ is cyclic, by definition of a cyclic group.

But here i do not understand the following: Why must $\langle a \rangle$ be a subgroup of $G$? For every single element $a$ in $G$, if $\langle a \rangle$ is a subgroup of $G$, then every group should have at least as many subgroups as the number of its elements. I would appreciate any help. Thanks

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$\langle a \rangle$ is a subgroup by definition. Look up the definitions before reading proofs using them. –  Martin Brandenburg Mar 17 '13 at 16:52
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And the converse holds if and only if the order of $G$ is prime. So actually, $G$ has no nontrivial subgroups if and only if $G\simeq \mathbb{Z}_p$ for some prime $p$. –  1015 Mar 17 '13 at 17:00
    
@Martin Bradenburg, yes the definition says it but i thought it doesn't need to be like that, but i was confused now i understood why. –  bigO Mar 17 '13 at 17:14
    
@julien, thanks –  bigO Mar 17 '13 at 17:15

3 Answers 3

up vote 6 down vote accepted

Given $a\in G$, $\langle a\rangle$ is defined to be the smallest subgroup of $G$ containing $a$ as an element. Knowing that $\langle a\rangle=\langle b\rangle$ isn't enough to conclude that $a=b$, though. For example, we always have $\langle a\rangle=\langle a^{-1}\rangle,$ no matter the order of $a$.

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Thanks for the answer, i see now. –  bigO Mar 17 '13 at 17:14

Every element of a group $G$ does indeed generate a subgroup. But the subgroups generated by different elements are not necessarily equal.

If $a\in G$, $\langle a \rangle$ is the smallest subgroup containing $a$, and it is by definition, a cyclic subgroup. It's elements do indeed form cyclic subgroups of \langle a \rangle as well, but note that any given element may be in a number of subgroups.

That is, subgroups need not be disjoint, in fact, they can't be, since at the very least $e$ is an element in every subgroup, by the definition of a subgroup. So an element can indeed be in more than one subgroup.

Hint

Look at $\mathbb Z$ and compare some of its subgroups: $2\mathbb Z$, $4\mathbb Z$, $8\mathbb Z$, etc. We have $8\mathbb Z \leq 4\mathbb Z \leq 2\mathbb Z \leq \mathbb Z$

Note, e.g., $2\in \{2n\mathbb Z: n \in \mathbb N, n\geq 1\} \leq \mathbb Z$. Note that $8 = 4(2n) \in 2n\mathbb Z$, but $8$ is also in $\mathbb 4n\mathbb Z$ and it generates $8n\mathbb Z$.

For a finite example, look at $\mathbb Z_{4} = \langle 1 \rangle = \langle 3 \rangle$. While $2 \in \langle 1 \rangle,\;\; \langle 2 \rangle \neq \langle 1 \rangle.$ But $\langle 2 \rangle = \{0, 2\} \leq \langle 1 \rangle.$ And clearly, $\langle 0 \rangle = \{0\} \leq \langle 2 \rangle \leq \langle 1 \rangle$

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Thanks for the answer, i see now. –  bigO Mar 17 '13 at 17:13

You mixed the one-element set $\{a\}$ and and the set $<a>$ of all powers of an element $a$.

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Thanks for the answer, i see now. –  bigO Mar 17 '13 at 17:15

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