free bases S and T are given, then the intersection of M and N is either the free subgroup P generated by the intersection of S and T or the free product of P with an infinite cyclic group. &lt;/p&gt; &lt;p class=“noindent”&gt; The main result of this article is that if M and N are Magnus subgroups (not necessarily distinct) of a one-relator group G and g and h are elements of G, then either the intersection of gMg&lt;sup&gt;-1&lt;/sup&gt; and hNh&lt;sup&gt;-1&lt;/sup&gt; is cyclic (and possibly trivial), or gh&lt;sup&gt;-1&lt;/sup&gt; is an element of NM in which case the intersection is a conjugate of the intersection of M and N. &lt;/p&gt;

Introduction {#sec1}
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A Magnus subgroup of a one-relator group $G=\langle X : r=1 \rangle$, where $r$ is cyclically reduced, is a subgroup generated by a Magnus subset $S$ of $X$, ie a subset $S$ which omits a generator explicitly occurring in the relator $r$. By the Freiheitssatz of Magnus (see for example page 104 or page 198 of Lyndon and Schupp [@LS]), any such subgroup is free with the given subset as basis.

The classical proof of many theorems on one-relator groups is by induction on the length of the 