Counting the Elements of a Finite Group [closed]

Let $G = \mathbb Z_6 \times \mathbb Z_4$, and find $[G:H]$ for:

a) $H = \{0\} \times \mathbb Z_4$

b) $H = \langle 2\rangle \times \langle 2\rangle$

• This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. Commented Oct 24, 2013 at 21:47

Determine the order of $G: \,|G|$, and the orders of $H$ in each case; then you can easily calculate the index of $H$ in $G$: $$[G:H] = \dfrac{|G|}{|H|}$$

Now, since the order of a group is the number of elements it contains, and the order of the direct product of two sets is given by the product of the orders: $$|G| = |\mathbb Z_6\times \mathbb Z_4| = |\{0, 1, 2, 3, 4, 5\}| \times |\{0, 1, 2, 3\}| = 6 \times 4 = 24$$ and so there are $24$ ordered pairs in $G$.

As for the number of elements of $H$, given

a. $H = \{0\} \times Z_4 = \{(0, 0), (0, 1), (0, 2), (0, 3)\}$. So the order of $H$ in this case is equal to $4 = 1\times 4$.

b. $H = \langle 2\rangle \times \langle 2\rangle = \{(0, 0), (0, 1), (1, 0), (1, 1)\}$. So the order of $H$ in this case is also equal to $4$.

• @Amzoti - Thanks for the support! ;-) Commented Oct 25, 2013 at 13:12

Relevant facts: $$|\mathbb{Z}_n|=n\\ |G\times G'|=|G|\cdot |G'| \\ [G\: :\: H]\cdot|H|=|G|\quad\mbox{(Lagrange's Theorem)}$$