What are the elements of the group $15 \mathbb Z/ 15 \mathbb Z$?

Question

My question is: What are the elements of the group $15 \mathbb Z/ 15 \mathbb Z$?

I am confused between if it is the 0 coset or the 1 coset. I want to conclude that $$12 \mathbb Z/ 24 \mathbb Z \times 15 \mathbb Z/ 15 \mathbb Z \times 25 \mathbb Z/ 50 \mathbb Z \cong \mathbb Z/ 2 \mathbb Z \times \mathbb Z/ 2 \mathbb Z.$$

Could someone explain this to me please?

Answer 1

Your confusion between being a $0$ coset or a $1$ coset is likely because you are mixing between multiplicative and additive notation. Since we are thinking of these in additive notation, $15\Bbb Z/15\Bbb Z\cong (\{0\},+)\cong (\{1\},\ast)$.

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A potential hazard one might run into when talking about these things was discussed in the comments. It may be wise to denote the identity element of a group $G$ by $e_G$. Inside a group $(G,\ast)$, one always has $(G,\ast)$ as a normal subgroup, and in particular there is a well defined quotient group $G/G$ with a single coset, which is represented by any element in $G$.

Recall that given a group $G$, and any subgroup $H\subseteq G$, one can talk about left cosets, represented by the elements of $G$. I.e. $gH =\{gh\mid h\in H\}$. The collection of these cosets will form a group if and only if $H$ is a normal subgroup of $G$.

One should avoid referring to a $0$ coset or a $1$ coset, and instead refer to a trivial coset, or the $e_G$ coset, just in case there are elements $0,1$ in the underlying set that one could believe you are referring to.

Furthermore, above, I used what is standard, namely a multiplicative notation. However, in the case of using additive notation. If we have subgroup $(H,+)$ inside of $(G,+)$, then a left coset of $H$ in $G$ is typically denoted $g + H =\{ g + h\mid h\in H\}$.

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In the case of $15\Bbb Z$, let us write $H= 15\Bbb Z$ for the (non-proper) subgroup $15\Bbb Z\subseteq 15\Bbb Z$, just for clarity. An element of $15\Bbb Z$ is of the form $15k$ for some $k\in \Bbb Z$. Each such element corresponds to a (say, left) coset $$15k + H=\{\dots, 15k-15, 15k, 15k+15,\dots\}$$ You can see for yourself that regardless of $k\in\Bbb Z$ that this defines the same set (simply note that $15k+15l = 15(k+l)$ for any $l\in \Bbb Z$), i.e. each choice of representative defines the same element. In particular, each coset is equivalent to the coset represented by the identity element $e_{15\Bbb Z}+15\Bbb Z$, i.e. the coset represented by the integer $0$.

Answer 2

Strictly speaking, the element of $15\Bbb Z/15\Bbb Z$ is the single coset

$$a+15\Bbb Z=15\Bbb Z,$$

where the equality holds for each $a\in 15\Bbb Z$.

This is because, for any quotient group $G/H$, its elements are of the form

$$g*H,$$

where $g\in G$ and $*$ is the operation of the group $G$, and the equality

$$g*H=k*H$$

holds exactly when $g*k^{-1}\in H$.