# Is $G/\langle x\rangle \cap \langle x \rangle=e$ given the following subgroups have order 11 and 7

I was trying to prove a question in group theory and if I assumed that $$G/\langle x\rangle \cap \langle x \rangle=e$$ it will solve my question. I believe this is true because I know that the order is as follows, $$|G/\langle x\rangle|=11$$ and $$|\langle x \rangle|=7$$. This means that I have two cyclic groups because both are prime and it then follows that their intersection can only be the identity element. But, the identity element of $$G/\langle x\rangle$$ is $$\langle x \rangle$$ so I am unsure if I am making a big mistake.

• $G/\langle x\rangle$ is a group whose underlying set is a set of cosets of $G$; $\langle x\rangle$ is a subset of $G$. They don’t even “live” in the same set, so asking about their intersection is nonsense. – Arturo Magidin Dec 18 '18 at 22:15
• @C Monsour: Your “correction” cannot be what was intended; for $\langle x\rangle$ is precisely the identity element of $/\langle x\rangle$; So you are basically rewriting it to ask if $G/\langle x\rangle \cap \{e_{G/\langle x\rangle}\} = \{e_{G/\langle x\rangle}\}$ (what you wrote also does not make sense because the left hand side is a subset, but the right hand side is an element). I suspect that what is really happening is that the OP is rather confused, and your “correction” neither clarifies, nor helps the OP recognize or correct the confusion. – Arturo Magidin Dec 20 '18 at 18:45
• I’ve rolled it back, because if we’re going to have nonsense, let’s have the OP’s nonsense, not someone else’s guess about what the nonsense ought to be. – Arturo Magidin Dec 20 '18 at 18:46

Actually $$G / \langle x \rangle \cap \langle x \rangle = \emptyset$$ since $$G / \langle x \rangle$$ has elements that are cosets and $$\langle x \rangle$$ contains elements of $$G$$.
• Perhaps -- unless one of the elements of $G$ just happens to be a set of other elements of $G$ that just happens to constitute a coset ... – Henning Makholm Dec 20 '18 at 18:52