3\mathbb{Z}$ contains free nonabelian subgroups?

Asked Feb 15 '25 at 12:34

Active Feb 15 '25 at 14:00

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The free product $ \mathbb{Z}/2\mathbb{Z} * \mathbb{Z}/3\mathbb{Z}$ contains free nonabelian subgroups.

Let's see presentation of a groups:

$ \mathbb{Z}/2 = \langle a \mid a^2 = e \rangle$

$ \mathbb{Z}/3 = \langle b \mid b^3 = e \rangle$

By the definition of a free product:

$G = \mathbb{Z}/2 * \mathbb{Z}/3 = \langle a,b \mid a^2 = b^3 = e \rangle$

I have a hypothesis that a ping-pong lemma can be used here, but I'm not sure if it is. I would be very grateful if you could give me any hints on how to do this.

edited Feb 15 '25 at 13:32

asked Feb 15 '25 at 12:34

Miganyshi

Not sure about that, but have you looked at the subgroup $\langle ab, ab^2 \rangle$? – Didier Feb 15 '25 at 12:48
Duplicate of https://math.stackexchange.com/questions/1951142/free-group-of-rank-2-inside-mathbbz-2-mathbbz-mathbbz-3-mathbbz but there the people have decided as usual to abuse the comment section to give answers, hence this cannot be mark as a duplicate target (as of now). – Martin Brandenburg Feb 15 '25 at 12:51
@Didier This is just the whole group. – Martin Brandenburg Feb 15 '25 at 12:55
@MartinBrandenburg Right, my bad. Ooops – Didier Feb 15 '25 at 12:56
2

See this question and this one – lulu Feb 15 '25 at 12:57
This can be done very explicitly using the Kurosh subgroup theorem, see: https://qchu.wordpress.com/2015/11/29/drawing-subgroups-of-the-modular-group/ – Qiaochu Yuan Feb 15 '25 at 13:00

Why free product $ \mathbb{Z}/2\mathbb{Z} * \mathbb{Z}/3\mathbb{Z}$ contains free nonabelian subgroups?

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