The cardinality of the minimal generating set of $G = \mathbb{Z}_2^m$ is $\log_2 |G|$

Question

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It is proven there that the minimum size of a generating set for a finite group at most $\log_2 |G|$?

In the answer, it is noted without a proof that:

The cardinality of the minimal generating set of $G = \mathbb{Z}_2^m$ is $\log_2 |G|$

Can anyone explain why is that true?

Sntn · Answer 1 · 2019-03-10T10:12:13.267

1

Hint: Think of $\mathbb{Z}_2^m$ as an $m$-dimensional vector space over $\mathbb{Z}_2$. Then, a minimum generating set really is a basis.

edited Mar 10 '19 at 10:12

answered Mar 10 '19 at 09:59

Sntn

How does that help me show the result? – Um Shmum Mar 10 '19 at 10:05
@UmShmum The minimum cardinality for that minimal generator would then be $m$ and the answer you referred showed that the maximum value is $m$ too. So, it must be equal to $m$ then. – Sntn Mar 10 '19 at 10:06
How do you show that the dimension of the basis is greater than $m$? – Um Shmum Mar 10 '19 at 10:09
@UmShmum What do you mean? Since this is an $m$ dimensional vector space, the cardinality of the basis is $m$. – Sntn Mar 10 '19 at 10:11
Sorry, I meant how do you show that this is an $m$-dimensional vector space over $\mathbb{Z}_2$? – Um Shmum Mar 10 '19 at 10:39
@UmShmum Because the set ${(1,0,...,0),(0,1,...,0),...,(0,0,...,1)}$ is a basis. – Sntn Mar 10 '19 at 10:58
Let us continue this discussion in chat. – Um Shmum Mar 10 '19 at 11:05
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Alternatively, since the group is abelian with exponent $2$, you could show that a subset of size $k$ generates a subgroup of order at most $2^k$. Hence to generate the whole group we must have $k \ge m$. – Derek Holt Mar 10 '19 at 12:01

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